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 takes 0.2 sec to travel the height of the window, which is 2 m.

 Then the speed of the ball at that time was

$v = \frac{d}{t}$

$v = \frac{2}{0.20}$

$v = 10\ \frac{m}{s}$

We know that:

$v = v_0 -gt$

Where $v_0$ is the initial velocity.

So:

$10 = v_0 -9.8(2)$

$10 + 9.8(2) = v_0$

$v_0 = 29.6\ m / s$

Then we have the equation for the position as a function of time.

$r = r_0 + v_0t - \frac{1}{2}gt^2$

Where

$r_0$ = initial position

r = position as a function of time

$v_0$ = initial velocity

g = acceleration of gravity

t = time.

If the ball is thrown from the ground then:

$r_0$ = 0 m

We want to find now the distance between the window and the ground.

When the ball reaches the bottom of the window t = 1.8s

So:

$r = 0 + 29.6(1.8) - 0.5(9.8)(1.8)^2$

$r = 37,404$ m

The window is 37,404 m high

Finally, to know how high the ball rises we must know at what moment the vertical velocity of the ball is zero.

$v = v_0 -gt\\\\0 = v_0 -gt\\\\gt = v_0\\\\t = \frac{29.6}{9.8}$

$t = 3.02\ s$

Now we replace t in the position equation

$r = 0 + 29.6(3.02) -0.5(9.8)(3.02)^2$

$r = 44.70\ m$

The ball reaches up to 44.70 m in height.