What Are the Unknowns in This Free Fall Ball Problem?

AI Thread Summary
The discussion focuses on solving a free fall ball problem involving a ball thrown upward with a speed of 32.3 m/s at half its maximum height. Key points include the use of constant acceleration equations to determine maximum height, velocity after 2 seconds, height after 2 seconds, and acceleration at maximum height. There is confusion regarding the application of equations for different segments of the ball's flight, particularly when transitioning from half height to maximum height. Participants emphasize the importance of clearly distinguishing between the first and second halves of the ball's trajectory to avoid errors in calculations. The conversation highlights the need for careful analysis of initial conditions and the correct application of kinematic equations.
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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.
 
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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.
 
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?
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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