I do not understand the answer that was given to me on a post dated Nov. 23.

Please help explain my questions that appear at the end of this post.
Bart
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Posted by Bart on Wednesday, November 23, 2011 at 6:52am.

A baseball is hit with a speed of 27 meters per second at an angle of 45 degrees. It lands on the flat roof of a 13 meter tall building. If the ball was hit when it was 1 meter above the ground, what horizontal distance does it travel before it lands on the building?

A long jumper leaves the ground at 45 degrees above the horizontal and lands 8 meters away.
What is his "takeoff" speed?

A long jumper is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10 meters away horizontally, and 2.5 meters vertically below. If he jumps from the edge of the left bank at a 45 degree angle with the speed calculated in question.
How long or short from the opposite bank will he land?
• physics - drwls, Wednesday, November 23, 2011 at 7:18am
#2: 8.0 m = (Vo^2/g)sin(2A) = Vo^2/g
Solve for Vo

For #1, Vx = Voy = 27/sqrt2 = 19.1 m/s

Y = 1 + Voy*t -(g/2)t^2 = 13
Solve for t. Then, use
X = Vx*t for the horizontal distance

For #3, use what you have learned and try it yourself. I assume you are supposed to use the takeoff speed calculated in the previous question.

For #2 What formula did you use when writing: (Vo^g)sin (2A)? Where did the 2 in the 2A come from?

For #1 Where did the 27/sqrt2 come from?

For #3 I still need help doing this one. I am not yet at the point where I can do it myself.

1. Vo = 27 m/s @ 45 Deg.

Xo = 27cos45 = 19.09 m/s.
Yo = 27sin45 = 19.09 m/s.

Vf = Vo + gt,
t = (Yf - Yo) / g ,
t = Tr = (0 - 19.09 / -9.8 = 1.95 s. =
Rise time = Time to reach max. ht.

hmax = Yo*t + 0.5g*t^2,
hmax = 19.09*1.95 - 4.9(1.95)^2 = 18.59 m.

d=Vo*t + 0.5t^2 = 18.59 - 13 = 5.59 m.
0 + 4.9t^2 = 5.59,
t^2 = 1.14,
t = Tf = 1.07 s. = Fall time = Time to
fall to roof.

T = Tr + Tf = 1.95 + 1.07 = 3.02 s. =
Time in flight.

Dx = Xo*T = 19.09 * 3.02 = 57.7 m. Hor
dist. traveled.

2. Dx = Vo^2*sin(2A)/g = 8 m.
Vo^2*sin(90) / 9.8 = 8,
Vo^2 / 9.8 = 8,
Vo^2 = 78.4,
Vo - 8.85 m/s. = Inital velocity.

3. Dx = (8.85)^2sin(90)/9.8 = 8 m.
d = 10-8 = 2 m. short.

Let's break down the questions and the explanations given by "drwls" in the post.

For question #2, drwls used the formula "8.0 m = (Vo^2 / g) * sin(2A)" to solve for the takeoff speed (Vo).

In this formula:
- "Vo" represents the takeoff speed, which is what we are trying to find.
- "g" represents the acceleration due to gravity.
- "A" represents the angle at which the long jumper leaves the ground, which is 45 degrees in this case.
- The sin(2A) term accounts for the fact that the jumper's motion can be split into a vertical component and a horizontal component, with the takeoff angle being double the angle measured from the horizontal.

Now, let's move on to question #1. To find the horizontal distance traveled before the ball lands on the building, drwls used the value of "27/sqrt2" for both the horizontal velocity (Vx) and vertical velocity (Voy) components.

In this case:
- "27" represents the initial speed of the ball.
- "sqrt2" is the square root of 2, which appears because the ball is launched at a 45-degree angle, and the horizontal and vertical velocities are equal.

To find the time of flight (t), drwls used the equation "Y = 1 + Voy * t - (g/2) * t^2 = 13" and solved for "t". "Y" represents the initial height of the ball, which is 1 meter above the ground, and "13" is the height of the roof.

To find the horizontal distance (X), drwls used the equation "X = Vx * t", which simply represents the distance traveled in the horizontal direction during the time "t" obtained earlier.

Finally, for question #3, drwls suggested that you use what you have learned from the previous questions and try to solve it yourself. However, you mentioned that you still need help with this question. Could you provide more details about what specifically you are struggling with in question #3?

For #2, the formula used is:

8.0m = (Vo^2/g)sin(2A) = Vo^2/g

The formula used here is derived from the equations of projectile motion. The symbol "Vo" represents the initial velocity of the long jumper, "g" represents the acceleration due to gravity, and "A" represents the launch angle (which is given as 45 degrees in the problem). The 2 in "2A" comes from a trigonometric identity that relates the sine of an angle to the sine of twice that angle.

For #1, the value of 27/sqrt2 comes from the components of the initial velocity of the baseball. When an object is launched at an angle, you can break down its initial velocity into horizontal and vertical components using trigonometry. In this case, the initial velocity of the baseball is given as 27 m/s at an angle of 45 degrees. The horizontal component (Vx) of this velocity can be found using the formula Vx = Vo*cos(A), where Vo is the initial velocity magnitude and A is the launch angle. Since cos(45 degrees) = sqrt2/2, we get Vx = 27/sqrt2.

For #3, let's assume you have already calculated the takeoff speed (Vo) in the previous question. To find out how far the long jumper will land from the opposite bank, you need to calculate the horizontal distance (X) traveled using the calculated takeoff speed.

Using the equation for horizontal distance, X = Vx*t, where Vx is the horizontal component of the takeoff speed and t is the time of flight.

Since the jumper is launching at a 45-degree angle, the horizontal and vertical components of the takeoff speed will be equal (Vx = Voy = Vo/sqrt2).

To calculate the time of flight(T), you can use the equation Y = 0 + Voy*T - (g/2)*T^2, where Y is the vertical distance (2.5m) and Voy is the vertical component of the takeoff speed.

Solving this equation will give you the time of flight (T). Substituting the value of T in the equation for horizontal distance, X = Vx*T, will give you the answer for question #3.