Question

A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance.

a. Find the time at which the two balls collide. Express your answer in terms of the variables H, v0, and appropriate constants.
b. Find the value of H in terms of v0 and g so that at the instant when the balls collide, the first ball is at the highest point of its motion. Express your answer in terms of the variables v0 and g.

Answer 1

a) t = H/v0

b) H = -(v0)²/g

Step-by-step explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g