Friday

May 6, 2016
Total # Posts: 620

**Physics repost**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 8, 2007*

**Physics**

Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court? What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???
*November 8, 2007*

**Physics**

Ok so I have the initial speed and the max height. Now it asks "At what initial speed must the ball be hit so that it lands directly on the opponent's back line?" Do I use a process similar to that of the first question?
*November 8, 2007*

**Physics**

Now it asks for the max. height. above the court reached by the ball. Do I use the equation Vf^2 = vi^2 + 2(g)(delta y) ?
*November 8, 2007*

**Physics**

A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground...
*November 8, 2007*

**Physics**

This is probably really obvious, but I'm still not undertanding why you have to subtract one.
*November 7, 2007*

**Physics**

How did you come up with 39.96 Vh^2?
*November 7, 2007*

**Physics**

The launching speed of a certain projectile is 6.4 times the speed it has at its maximum height. Calculate the elevation angle at launching. I have noo idea how to start this one. Please help!
*November 7, 2007*

**Physics**

The soccer goal is 23.05 m in front of a soccer player. She kicks the ball giving it a speed of 17.97 m/s at an angle of 25.83 degrees from the horizontal. If the goalie is standing exactly in front of the net, find the speed of the ball just as it reaches the goalie. Do I ...
*November 6, 2007*

**Physics**

Thanks!
*November 6, 2007*

**Physics**

A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 46.7° above the horizontal, and with a speed v = 28.4 m/s. Assuming that air friction ...
*November 6, 2007*

**Physics**

A stone is aimed at a cliff of height h with an initial speed of 62 m/s directed 45° above the horizontal, as shown in the Figure below. The stone strikes at A, 6.56 s after launching. What is the height of the cliff? Can someone please tell me which equation(s) I need to ...
*November 6, 2007*

**Physics**

A helicopter is flying in a straight line over a level field at a constant speed of 21.7 m/s and at a constant altitude of 11.5 m. A package is ejected horizontally from the helicopter with an initial speed of 21.6 m/s relative to the helicopter, and a direction opposite to ...
*November 4, 2007*

**Physics**

All right n/m, I finally got the answer. :)
*November 4, 2007*

**Physics**

All right. So this is what my equation looks like so far. I need to solve for Vx: 25Vx^2 = 313.786 + Vx^2 Correct?
*November 4, 2007*

**Physics**

And drwls...where is 0.2041 coming from?
*November 4, 2007*

**Physics**

But my number was correct, right? 17.714 m/s?
*November 4, 2007*

**Physics**

A ball is thrown horizontally from a height of 16.01 m and hits the ground with a speed that is 5.0 times its initial speed. What was the initial speed? Ok so I got a vertical final velocity of 17.714 m/s. Now I need to find the horizontal speed, correct? Which equation can I ...
*November 4, 2007*

**Physics**

Ok makes sense. The only part I'm a little confused about is finding the horizontal speed...which equation is that?
*November 4, 2007*

**Physics**

A ball is thrown horizontally from a height of 16.01 m and hits the ground with a speed that is 5.0 times its initial speed. What was the initial speed? This question really boggles my brain. I have no idea how to solve this, given only the height. Could someone please steer ...
*November 4, 2007*

**Physics**

Ahh ok I got it now. Thanks. :)
*November 4, 2007*

**Physics**

Once I then have the initial vel., how can I go about getting the horizintal distance? Don't I need time?
*November 4, 2007*

**Physics**

A small steel ball bearing with a mass of 24 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.23 m. Calculate the horizontal distance the ball will travel if the same spring is aimed 27° from the ...
*November 4, 2007*

**Physics**

15.6 m/s. I understand now, thanks a bunch.
*November 3, 2007*

**Physics**

A stone thrown horizontally from a height of 7.8 m hits the ground at a distance of 12.0 m. Calculate the speed of the ball as it hits the ground. I was able to find the initial speed of the ball, which is 9.52 m/s. However, I'm unsure about whether I need to use this ...
*November 3, 2007*

**Physics**

A snowball is launched horizontally from the top of a building at v = 15 m/s. If it lands d = 44.5 meters from the bottom, how high (in m) was the building? For this problem, do I need to solve for any other velocities? I'm a little stuck...I'm not quite sure how to ...
*November 2, 2007*

**Physics**

THANK YOU!
*November 2, 2007*

**Physics**

Salmon, swimming up the Fraser river to their spawning grounds, leap over all sorts of obstacles. The unofficial salmon-altitude record is an amazing 3.58 m jump. Assuming the fish took off at 45.0°, what was its speed on emerging from the water? Ignore friction. This ...
*November 2, 2007*

**Physics**

Thanks, I got it now. :)
*November 2, 2007*

**Physics**

A rescue plane has to drop supplies to a group of castaways who are stranded on a deserted island. The plane is flying as 150 m/s at a level altitude of 1200 m. How far ahead of the landing zone should the plane release the supplies? I'm not even sure how to start this one...
*November 2, 2007*

**Physics**

At the entrance channel of a harbor, the tidal current has a velocity of 4.94 km/hr in a direction 23.2° south of east. Suppose a ship caught in this current has a speed of 15.6 km/hr relative to the water. If the helmsman keeps the bow of the ship aimed north, what will ...
*October 18, 2007*

**Physics**

At the entrance channel of a harbor, the tidal current has a velocity of 4.94 km/hr in a direction 23.2° south of east. Suppose a ship caught in this current has a speed of 15.6 km/hr relative to the water. If the helmsman keeps the bow of the ship aimed north, what will ...
*October 18, 2007*

**Physics**

*not sure
*October 16, 2007*

**Physics**

At the entrance channel of a harbor, the tidal current has a velocity of 4.94 km/hr in a direction 23.2° south of east. Suppose a ship caught in this current has a speed of 15.6 km/hr relative to the water. If the helmsman keeps the bow of the ship aimed north, what will ...
*October 16, 2007*

**Physics**

Great I got it, thanks. :)
*October 16, 2007*

**Physics**

I am given a velocity versus time graph for a runner. For it, I had to figure out the average velocity for the entire 16 s interval (for this, I got 2 m/s). Now they want me to find the displacement of the runner in the first 15.4 s. I know that it must be somewhere around 30...
*October 16, 2007*

**Physics**

A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 12.6 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B. I am still stuck on this problem, mainly because I can't seem to get v/2. ...
*October 15, 2007*

**Physics**

See, I use your method without rounding and now come up with 0.0852 s. So...now I'm a little stuck on which answer is the correct one.
*October 10, 2007*

**Physics**

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump? Ok...after this I swear I'm done! I just need someone to check my final answer, which is 0.0858 s.
*October 10, 2007*

**Physics**

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump? I need someone to check my work. I got about 3.797 as the inital velocity. I plugged this into "x = vt + 1/2at...
*October 10, 2007*

**Physics**

But don't I need time in order to find the initial velocity?
*October 10, 2007*

**Physics**

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump? I've been told that I need to use the height of the jump to solve for V. Then I need to use that V in the ...
*October 10, 2007*

**Physics**

How can I solve for V if I don't yet have the time?
*October 9, 2007*

**Physics**

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the bottom 15.4 cm of the jump? I think I have this right, but I'd like someone to make sure. First I have to find the total time by plugging ...
*October 9, 2007*

**Physics**

When finding the times, do I have to switch the cm to m?
*October 9, 2007*

**Physics**

Ok n/m, I got the answer. However, now it asks for the time in the bottom 15.4 cm. Do I have to use my time from the first problem to figure it out?
*October 9, 2007*

**Physics**

I found the time. Now I just have to double it, and that will be the time he spent in the top 15.4 cm?
*October 9, 2007*

**Physics**

To find the time, would I have to use the formula "x = vt + 1/2at^2?"
*October 9, 2007*

**Physics**

A basketball player, standing near the basket to grab a rebound, jumps 73.6 cm vertically. How much time does the player spend in the top 15.4 cm of his jump? I never know which formula to use for these problems! Any help will be greatly appreciated.
*October 9, 2007*

**Physics**

Never mind! The mass has nothing do with the problem. I got it. :)
*October 9, 2007*

**Physics**

It takes 2.5 s for a small ball released from rest from a tall building to reach the ground. The mass of the ball is 0.05 kg. Calculate the height from which the ball is released. Ok so I have the time, initial velocity, and the acceleration. But the mass totally throws me off...
*October 9, 2007*

**Physics**

Ohhh ok I see my mistake! Thanks for your patience. :)
*October 8, 2007*

**Physics**

Right. I plug in "1/2 x 3.05 x (2.58^2") and come out with 10.15 m. But that's obviously not correct. I know I'm doing something wrong, but I'm just not sure what. I'm sorry...this just isn't making much sense yet.
*October 8, 2007*

**Physics**

Which equation am I using for distance?
*October 8, 2007*

**Physics**

Ok, the time I got was 2.624 s. Now what?
*October 8, 2007*

**Physics**

What exactly do you mean by "set them equal?"
*October 8, 2007*

**Physics**

At the instant a traffic light turns green, a car starts moving with a constant acceleration a of 3.05 m/s^2. At the same time a truck, traveling with a constant speed of 10.15 m/s, overtakes and passes the car. How far beyond the traffic light will the car overtake the truck...
*October 8, 2007*

**Physics**

Now, when I'm solving for time, do I not yet know what v/2 is? Like, it would just be "t = v/2 x -9.8?" I'm so sorry...this is just not making any sense to me yet.
*October 7, 2007*

**Physics**

A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 12.6 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B. I was told the equation to this was "h=1/2(v2)^2/g." However, I’...
*October 7, 2007*

**Physics**

So, am I supposed to figure out v2? And what is g, in this case?
*October 5, 2007*

**Physics**

A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 12.6 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B. What equation can I use to get the maximum height?
*October 5, 2007*

**Sociology**

What exactly is "cultural purity?" I think I understand the concept, but I'm having a hard time putting it into words.
*October 4, 2007*

**Physics**

THANK YOU!
*October 4, 2007*

**Physics**

A golf ball released from a height of 1.80 m above a concrete floor, bounces back to a height of 1.06 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor? Quidditch told ...
*October 4, 2007*

**Physics**

What information must I use to get the velocities?
*October 4, 2007*

**Physics**

A golf ball released from a height of 1.80 m above a concrete floor, bounces back to a height of 1.06 m. If the ball is in contact with the floor for 4.62 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor? Do I need to ...
*October 4, 2007*

**Physics**

Ok great, I get it now. Thanks!
*October 4, 2007*

**Physics**

**Oops sorry, my sister was using this site last night for Chemistry.
*October 4, 2007*

**Physics**

-The brakes on your automobile are capable of creating acceleration of 19 ft/s2. If you are going 76 mi/hr and suddenly see a state trooper, what is the minimum time required to get your car to 55 mi/hr? For this answer, I got 1.62 s. -How far do you travel while returning to ...
*October 4, 2007*

**Physics**

THANKS! :)
*October 3, 2007*

**Physics**

Ok, so I came up with 16.37 s as my answer for the time. So I do I simply plug that number into the equation 'distance=1/2 x a x t^2' and that is my runway length?
*October 3, 2007*

**Physics**

43 knots is 22.097 m/s. So could I now do 22.097 divided by 1.35?
*October 3, 2007*

**Physics**

To get how many seconds, would I have to take the 43 knots and divide it by 1.5?
*October 3, 2007*

**Physics**

I'm sorry. There was a problem before it where I had to figure out the average acceleration, and that was 1.35 m/s^2. Which fomula can I use to get the length of the runway?
*October 3, 2007*

**Physics**

43 knots, right?
*October 3, 2007*

**Physics**

The airplane can take off when its airspeed (speed of the air flowing over the wing) is equal to 65 knots. What is the length of runway required for the plane to take off if there is a 22 knots head wind? The runway at the Tallahassee Regional Airport has a length of 8000 ft. ...
*October 3, 2007*

**Physics**

N/m, I needed to convert to knots to meters per second. I got the right answer. Thanks for the help, Quidditch. :)
*October 3, 2007*

**Physics**

I try this and I get 2.62 knots/s^2, but it says this answer is wrong. Is it?
*October 3, 2007*

**Physics**

A Cessna airplane can accelerate from rest to a ground speed of 78 knots in 29.8 s. What is the average acceleration of this airplane? I figured out that 78 knots is equal to 89.76 miles, 144.45 km, or 144,450 m. And I think the equation might be average=delta V over delta t. ...
*October 3, 2007*

**Math**

Again, I'm learning about trigonometric functions. I need to find the missing information for each triangle. I have the length of one leg as 24 and the opposite theta as 72. I need to figure out the other angles and side lengths, but I'm really not sure how to.
*October 2, 2007*

**Math**

Ok cool, I got that. Thanks a lot. However, this next one gives me the length of one leg (359) and the adjacent theta of 83. How exactly can I solve this?
*October 2, 2007*

**Math**

I'm learning about trigonometric functions, and my worksheet instructs me to find the "missing information" for each triangle. For the first one, they give me the length of the hypotenuse (45) and that theta = 35. How can I go about finding the missing info?
*October 2, 2007*

**Physics**

A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m. a) What is the max. height reached by the rocket? b) When does the rocket reach max. ...
*September 30, 2007*

**Physics**

Ok so I plugged in the numbers and got 42.1 for a) and -11.4 for b). However, my book says that I got them switched...that 11.4 s is a) and -42.1 m/s is b). How did this happen?
*September 25, 2007*

**Physics**

A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. a) Find the velocity with which the camera hits the ground. b) Find the time required for it to hit the ground. Ok so I'm not even sure of which equation...
*September 25, 2007*

**Physics**

**Oops, I meant vf+vi. Is that the right equation?
*September 16, 2007*

**Physics**

In order to get the time, would I have to do the vf-vi divided by 2, times the distance?? I guess I really do need a brush-up on algebra!
*September 16, 2007*

**Physics**

I have the equation 'change in distance = 1/2 x (vf+vi) x change in time (or delta t)' for uniform acceleration. However, I am totally lost when it comes to rearranging it. I need to rearrange it for inital speed (vi) and change in time (delta t). Can someone please help?
*September 16, 2007*

**Physics**

A bicyclist accelerates -0.87 m/s/s during a 3.8 s interval. What is the change in velocity of the bicyclist and the bicycle? For this problem, I used the formula 'change in v = a x t" and came up with -3.306 m/s as my answer. Is this right?
*September 13, 2007*

**Physics**

Thanks. :)
*September 13, 2007*

**Physics**

A freight train traveling with a velocity of -4.0 m/s begins backing into a train yard. If the train's average acceleration is -0.27 m/s/s, what is the train's velocity after 17 seconds? Can someone tell me which formula to use for this?
*September 13, 2007*

**Physics**

Ok good. Then, for part B...I got 76.8 for the average velocity. ...I sure hope this is correct...!
*September 10, 2007*

**Physics**

So for part A, I got 6.38 hours because I added 3.18, 2.8 (210 km/75) and .40 together. Is this correct?
*September 10, 2007*

**Physics**

A bus travels 280 km south along a straight path with an average velocity of 88 km/hr to the south. The bus stops for 24 min, then it travels 210 km south with an average of 75 km/hr to the south. A. How long does the total trip last? B. What is the average velocity for the ...
*September 10, 2007*

**Algebra**

Could you explain to me how to these type of problems? I checked in my book and it is still confusing. Evalute the piecewise function at the given value of the independent variable. 1. g(x)= {x^2+2 if x cannont equal 2, x+8 if x=2 Determine g(-5). 2. f(x) {-5x+4 if x <-3, ...
*August 30, 2007*

**Human Services Administrator**

I am having trouble finding information on Government regulations dealing with teen abstinence. The sites given to me by my instructor provided are not giving me the information I need to complete my paper. I have done a search on this but only limited results. You are ...
*July 29, 2007*

**Science**

Is there an experiment I could do to prove that soap is made up of fat and lye? Thank you for using the Jiskha Homework Help Forum. Here is a recipe for making soap: Recipes From Scratch Basic Recipes Soap Crafters Castile (Non-veggie) 26 ounces of Olive Oil 60 ounces of Lard ...
*July 17, 2007*

**Science**

I'v been looking all over the web, but I can't seem to come up with a good response to this: "Describe 2 ways in which maple trees are suited to survive through the winter." Any ideas? Maple trees drop their leaves and the sap stops running in cold weather. ...
*June 27, 2007*

**Chemistry**

These are all true or false questions, and I think some of my answers may be wrong. Could someone please check them for me? 1) All stoichiometry calculations are based on STP conditions. (True) 2. Given the amount of reactant, you must use coefficients from the balanced ...
*June 4, 2007*

**Chemistry 3**

How is it possible for me to tell which of these solutes will produce an acidic solution when dissolved in water? LiOH CH3CO2K H2SO3 KOH NaOH NH4NO3 NaOH NaOH can't be the only one... Molecules that produce OH^- when dissolved in water will be basic. Examples:LiOH, KOH, ...
*May 24, 2007*