Question

Carlos and Jeff are playing catch by the math building. Carlos is on the ground, and Jeff is on top of MSB, 49 feet off the ground.

a. With what initial velocity must Carlos toss the ball up so that Jeff can (just barely) catch it?
b. If Jeff then drops the ball off the roof, with what velocity will it hit the ground?

Can someone explain how to solve these problems?

Answer 1

You need a formula from physics for height as a function of time h(t) for a ball that is thrown up.

Answer 2

I never took physics so I have no idea what formula you are talking about...

Answer 3

h(t) = ut + 1/2gt^2
u is the initial velocity with which the ball is thrown up.
t is time
h is height at time t
g is acceleration due to gravity.

Answer 4

\[h(t) = -\frac{1}{2}gt^2 + ut + h_0\]
\(h_0\) is initial height. \(g = 32 \text{ ft/s}^2\) or \(g = 9.8 \text{ m/s}^2\)

Answer 5

"He barely catches it" implies the maximum height reached by the ball is 49 feet. Also, the velocity of the ball will be zero when it reaches the maximum height.

Answer 6

here \(h_0 = 0\) (probably not actually 0, but we have to assume that because it doesn't tell us otherwise).

Agree with @ranga's description of the constraints.

Answer 7

velocity will be \[v(t) = -gt + u\]if I'm remembering correctly.

Answer 8

She said she hasn't taken Physics. So I just gave her the formula without the initial height because the problem doesn't specify it. Usually they will give the formula but here they did not.

Answer 9

The maximum height can also be computed using energy conservation principle. 1/2mu^2 = mgh where u is the initial velocity and h is the maximum height reached = 49 feet.

Answer 10

\[h(t) = -\frac{1}{2}gt^2 + ut + h_0\]
To find the maximum height, we need to find dh/dt, equate it to 0, solve for t, and put t back in h(t) to find maximum h. It so happens dh/dt is also the velocity of the body, v(t), with respect to time, and equating it to zero tells us that the velocity will be 0 at maximum height, which makes sense, because if there is any velocity left the ball will continue to go up and would not have yet yet reached the maximum height.

Yes, @whpalmer. v(t) = dh/dt = u - gt