A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?
A) 17 m
B) 21 m
C) 18 m
D) 27 m
E) 31 m

Question

Quizplus · Accepted Answer

Let's first calculate the maximum height that the ball reaches before starting to fall back down. We can use the formula:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the maximum height we're looking for).

0^2 = 20^2 + 2(-9.8)s
s = 20.4 m

So the maximum height that the ball reaches is 20.4 m.

Now let's look at the stone. We want to find the time it takes for the stone to reach the height of 20.4 m. We can use the formula:

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

where s is the displacement (20.4 m), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time we're looking for.

20.4 = 24t + (1/2)(-9.8)t^2
t = 2.05 s

So it takes the stone 2.05 s to reach the height of 20.4 m.

During this time, the ball has been in the air for a total of 2 + 2.05 = 4.05 s. We can use the formula:

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

to find how far the ball has fallen during this time. We'll take upward to be positive, so a = -9.8 m/s^2.

s = 20(4.05) + (1/2)(-9.8)(4.05)^2
s = 82.4 m

So the ball has fallen 82.4 m during the 4.05 s that the stone was in the air.

Now we can add up the maximum height of the ball and the height that the ball has fallen to find the height of the ball when it passes the stone:

height = 20.4 m + 82.4 m = 102.8 m

Since the question is asking for the height above the release point, we subtract the initial height of the ball:

height above release point = 102.8 m - 20 m = 82.8 m

However, note that this height is above the release point of the ball and not the release point of the stone (which is also at 20 m). Therefore, we need to subtract the height that the stone has traveled during the 2.05 s it took to reach the height of the ball. We can use the same formula as before, but with u = 24 m/s and t = 2.05 s:

s = 24(2.05) + (1/2)(-9.8)(2.05)^2
s = 24.9 m

So the stone has traveled 24.9 m since it was released. Therefore, the height above the release point of the stone is:

height above release point = 82.8 m - 24.9 m = 17.9 m

Rounded to the nearest whole number, the answer is 18 m, which corresponds to choice (C).

A Ball Is Thrown Vertically Upward with an Initial Speed