Free fall ball height problem

  • #1

Homework Statement


A ball is thrown upward in such a way that its speed is 32.3 m/s when it is at half its maximum height. Find
(a) its maximum height,
(b) its velocity 2.0 s after it's thrown,
(c) its height 2.0 s after it's thrown, and
(d) its acceleration at its maximum height.

Homework Equations


Constant acceleration equations:
v = v0 + at
x - x0 = v0t + 1/2 (at2)
v2 = v02 + 2a(x - x0)
x - x0 = 1/2 (v0 + v)t
x - x0 = vt - 1/2 (at2)

where v= velocity, a = acceleration, t = time, v0 = initial velocity and x0 = initial position

The Attempt at a Solution



For (a), I thought the acceleration should be constant, since it is a free fall problem. At the ball's maximum height, v should equal 0. So I said:

0 = 32.3 m/s - (9.8 m/s2)(t seconds)
t = 3.30 seconds, which should be the time it takes the ball to go from the halfway point to its maximum height, assuming subbing in speed for velocity didn't make the equation explode. I'm having trouble getting anything else useful without knowing the initial velocity or some other piece of information.
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
If h is the height, then

v^2 - vo^2 = - 2*g*h/2 ...(1)

0 - vo^2 = - 2*g*h.........(2) Or gh = vo^2/2. Substitute this value in the eq.(1)

v^2 - vo^2 = vo^2/2

v is given. Solve for vo and proceed.
 
  • #3
I'm not sure I understand:

in equation 1, we're using h/2, so the ball is at half height.
then in equation 2, we're setting v = 0, indicating the ball is at its maximum height.

Since those two equations are partially solved for different values of x, we can't sub them into one another correctly, can we?
 
  • #4
Delphi51
Homework Helper
3,407
11
rl.bhat has given us a very nice method - much better than solving the quadratic distance formula for time as I did!

You are quite right, we must be very careful using a formula for two different parts of the flight. I would carefully write a heading for "second half of rise" and another for "whole rise". In the first part, you can use the 32.3 m/s as the initial velocity and h/2 as the distance. In the second part, the initial v is unknown and the distance is h.
 

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