Question

A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and passes the top of the window 0.20 s later. The window is 2.0 m high from top to bottom. What was the ball's initial velocity? The unit vector j^ is directed upward. How far is the bottom of the window from the launch position? How high does the ball rise above the launch position?

Answer 1

The ball's initial velocity is 19.6 m/s. The distance from the launch position to the bottom of the window is approximately 19.404 meters. The ball rises 19.6 meters above the launch position.

Step-by-step explanation:

To find the ball's initial velocity, we can first calculate the time it takes for the ball to reach the bottom and top of the window.

From the given information, we know that it takes 0.20 seconds for the ball to pass the top of the window and 1.8 seconds for it to pass the bottom of the window.

The window is 2.0 meters high from top to bottom.

Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration (which is due to gravity and is approximately -9.8 m/s^2), and t is the time, we can find the initial velocity at the bottom of the window:

v = u + at

v = 0 + (-9.8)(1.8)

v = -17.64 m/s

Since the velocity is negative, it means the ball is moving upward.

Now, we can find the initial velocity at the top of the window:

v = u + at

v = 0 + (-9.8)(2.0)

v = -19.6 m/s

The negative sign indicates that the velocity is upward.

Since the velocity changes by 19.6 m/s in 0.20 seconds, we can calculate the average acceleration:

a = (v - u) / t

a = (-19.6 - (-17.64)) / 0.20

a = -196 m/s^2

Now, we can use the formula v = u + at to find the initial velocity:

v = u + at

-19.6 = u + (-196)(0.20)

-19.6 = u - 39.2

u = -19.6 + 39.2

u = 19.6 m/s

Therefore, the ball's initial velocity is 19.6 m/s.

To find the distance from the launch position to the bottom of the window, we can use the formula d = ut + (1/2)at^2, where d is the distance, u is the initial velocity, t is the time, and a is the acceleration:

d = ut + (1/2)at^2

d = 19.6(1.8) + (1/2)(-9.8)(1.8)^2

d = 35.28 - 15.876

d = 19.404 m

Therefore, the distance from the launch position to the bottom of the window is approximately 19.404 meters.

To find how high the ball rises above the launch position, we can use the same formula, but only consider the time it takes for the ball to reach the top of the window:

d = ut + (1/2)at^2

d = 19.6(2.0) + (1/2)(-9.8)(2.0)^2

d = 39.2 - 19.6

d = 19.6 m

Therefore, the ball rises 19.6 meters above the launch position.