PLEASE HELP!
A person throws a rock horizontally off a cliff ata a speed of 9 m/s from a point 64.8 m above the ground. How long is the rock in the air?
13.2 s
26.4 s
3.64 s(my guess)
5.14 s
A package falls from an airplane in level flight at constant speed. If air resistance can be neglected, how does the motion of the package look to the pilot.
The package appears to travel in the shape of a parabola.
The package appears to fall straight downward.(my guess)
The package appears to travel backward.
The package appears to fall backward and downward.
A paintball is shot horizontally in the positive x direction. At time Trianglet after the ball is shot, it is 4 cm to the right and 4cm below it’s starting point. At time 2trianglet, what is the position of the ball relative to it’s starting point. Ignore air resistance.
12 cm to the right and 16 cm below
8cm to the right and 8 cm below
12 cm to the right and 8 cm below
8 cm to the right and 16 cm below(my guess)
A baseball pitcher throws a fastball horizontally at a speed of 42.0 m/s. Ignoring air resistance, how far does the ball drop between the pitcher’s mound and home plate, 60 ft 6in away?
0.945 m
2.15 m
1.02 m
0.382 m
A rescue plane spots a person floating in a lifeboat 55 m directly below and releases an emergency kit with a parachute. The package descends with a constant vertical acceleration of 6.91 m/s. If the horizontal speed of the plane was 70.6 m/s, how far is the package from the lifeboat when it hits the waves?
225 m
1.27 km
323 m
282 m
To solve these problems step-by-step, let's tackle each one separately:
1. How long is the rock in the air?
Given: Initial velocity (v0) = 9 m/s, Height (h) = 64.8 m
Since the rock is thrown horizontally, the vertical component of its velocity is 0 m/s.
Using the kinematic equation: h = v0t + (1/2)gt^2, where g is the acceleration due to gravity (9.8 m/s^2), and solving for t:
64.8 = 0*t + (1/2)*9.8*t^2
64.8 = 4.9t^2
t^2 = 64.8/4.9
t^2 = 13.2245
t ≈ 3.64 s
So, the rock is in the air for approximately 3.64 seconds. Therefore, the correct answer is 3.64 s.
2. How does the motion of the package look to the pilot?
Given: The package falls from an airplane in level flight at constant speed, and air resistance is neglected.
When air resistance is neglected, the motion of a falling object follows a parabolic trajectory. Therefore, the correct answer is "The package appears to travel in the shape of a parabola."
3. What is the position of the ball relative to its starting point at time 2trianglet?
Given: At time Trianglet, the ball is 4 cm to the right and 4 cm below its starting point.
Since the ball is shot horizontally, its horizontal velocity remains constant over time.
At time 2trianglet, the horizontal position of the ball would be twice the distance traveled in time Tringlet, which is 8 cm to the right.
Since there is no vertical acceleration, the ball will fall vertically and accelerate downwards due to gravity. At time 2trianglet, the ball will be 2*(4 cm) = 8 cm below its starting point.
Therefore, the correct answer is "8 cm to the right and 8 cm below."
4. How far does the ball drop between the pitcher’s mound and home plate?
Given: Horizontal velocity (v) = 42.0 m/s, Horizontal distance (d) = 60 ft 6 in (which converts to 18.44 m)
Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s.
Using the kinematic equation: d = v0t + (1/2)gt^2, where g is the acceleration due to gravity (9.8 m/s^2), and solving for d:
18.44 = 0*t + (1/2)*9.8*t^2
18.44 = 4.9t^2
t^2 = 18.44/4.9
t^2 = 3.76326
t ≈ 1.94 s
The ball drops vertically, so the distance it drops is given by: d = (1/2)gt^2
d = (1/2)*9.8*(1.94)^2
d ≈ 1.02 m
Therefore, the ball drops approximately 1.02 meters between the pitcher's mound and home plate. So, the correct answer is 1.02 m.
5. How far is the package from the lifeboat when it hits the waves?
Given: Vertical acceleration (a) = 6.91 m/s^2, Horizontal velocity (v) = 70.6 m/s
The vertical motion of the package is independent of its horizontal motion, so we can analyze them separately.
For the vertical motion, we can calculate the time it takes for the package to hit the waves using the equation: Δy = v0t + (1/2)at^2, where Δy is the vertical displacement (55 m), v0 is the initial vertical velocity (0 m/s), a is the vertical acceleration (6.91 m/s^2), and t is the time.
55 = 0*t + (1/2)*6.91*t^2
55 = 3.455t^2
t^2 = 55/3.455
t^2 ≈ 15.9225
t ≈ 3.99 s
For the horizontal motion, we can calculate the horizontal distance using the equation: d = vt, where d is the horizontal distance and t is the time.
d = 70.6 * 3.99 ≈ 281.594 m
Therefore, the package is approximately 282 m from the lifeboat when it hits the waves. Hence, the correct answer is 282 m.
To summarize:
1. How long is the rock in the air? The rock is in the air for approximately 3.64 seconds.
2. How does the motion of the package look to the pilot? The package appears to travel in the shape of a parabola.
3. What is the position of the ball relative to its starting point at time 2trianglet? The ball is 8 cm to the right and 8 cm below its starting point.
4. How far does the ball drop between the pitcher’s mound and home plate? The ball drops approximately 1.02 meters.
5. How far is the package from the lifeboat when it hits the waves? The package is approximately 282 m from the lifeboat when it hits the waves.
To solve each of these problems, we need to use the principles of kinematics, specifically the equations of motion in the x and y directions.
1. To find the time the rock is in the air, we can use the equation of motion in the y-direction:
y = y0 + v0y * t - 0.5 * g * t^2
where y0 is the initial height (64.8 m), v0y is the initial vertical velocity (0 m/s since the rock is thrown horizontally), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. We want to find the time it takes for the rock to hit the ground, so y = 0. Rearranging the equation, we get:
0 = 64.8 - 0.5 * 9.8 * t^2
Solving for t gives us t ≈ 3.64 s, so the correct answer is 3.64 s.
2. When a package falls from an airplane with negligible air resistance, it follows a projectile motion and appears to the pilot as if it is falling straight downward. The correct answer is "The package appears to fall straight downward."
3. To find the position of the paintball at time 2t relative to its starting point, we need to consider its horizontal and vertical motion separately. Since the paintball is shot horizontally, its position in the x-direction does not change with time. Thus, the horizontal position at time 2t is 4 cm to the right. In the y-direction, we can use the equation of motion:
y = y0 + v0y * t - 0.5 * g * t^2
At time t, the paintball is 4 cm below its starting point, so y = - 4 cm. Plugging in the values, we get:
-4 = 0 + 0 - 0.5 * 9.8 * (2t)^2
Simplifying gives us 4t^2 ≈ 0.408. Solving for t gives us t ≈ 0.8 s. So at time 2t (which is 2 * 0.8 s = 1.6 s), the paintball's position is 8 cm below its starting point. Since the horizontal position does not change, the correct answer is "8 cm to the right and 8 cm below."
4. To find how far the baseball drops between the pitcher's mound and home plate, we can use the equation of motion in the y-direction. Since the ball is thrown horizontally, its initial vertical velocity is 0. The distance it drops is given by:
y = y0 + v0y * t - 0.5 * g * t^2
where y0 is the initial height (0 m), v0y is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes to travel the horizontal distance. We need to convert the given horizontal distance from feet and inches to meters: 60 ft 6 in is approximately 18.44 m. We can use the equation for horizontal motion to find the time it takes to travel this distance:
x = v0x * t
where v0x is the initial horizontal velocity (42.0 m/s), x is the horizontal distance to be covered (18.44 m), and t is the time. Solving for t gives us t ≈ 0.438 s. Now we can plug this value of t into the equation for y to find the drop distance:
y = 0 + 0 - 0.5 * 9.8 * (0.438)^2 ≈ 0.945 m. So the correct answer is 0.945 m.
5. To find how far the package is from the lifeboat when it hits the waves, we need to analyze its vertical and horizontal motion separately. In the vertical direction, we can use the equation of motion:
y = y0 + v0y * t + 0.5 * a * t^2
where y0 is the initial height (55 m), v0y is the initial vertical velocity (0 m/s since the package is released from rest), a is the vertical acceleration (6.91 m/s^2), and t is the time. We want to find the time it takes for the package to hit the waves, so y = 0. Rearranging the equation, we get:
0 = 55 + 0 + 0.5 * 6.91 * t^2
Solving for t gives us t ≈ 3.18 s. Now we can use the equation for horizontal motion:
x = v0x * t
where v0x is the initial horizontal velocity (70.6 m/s), x is the horizontal distance to be covered, and t is the time. Plugging in the values, we get:
x = 70.6 * 3.18 ≈ 225 m. So the correct answer is 225 m.
Five questions at once? You would be better helped to make it one question per post. Few volunteer tutors have time to sit down and do all five at one sitting. I will do a couple, maybe that will help you on the others. We do much better critiquing your WORK, as it is one wonders if you are an answer grazer.
First one: time in air.
hf=hi+vi*t-4.8t^2 *vi is initial vertical speed*
0=64.8-4.8t^2
t= sqrt(64.8/4.8)= your guess is the best answer.
fourth one:
convert 60 ft 6in to m: 18.4m
time in air: 18.4m/42m/s=.44 sec
hf=hi-4.8t^2
hf=0-4.8(.44^2)=.93m fell