How do I find vertical height knowing only hang time?

AVReidy

1. The problem statement, all variables and given/known data


A basketball player achieves a hang time of 0.843 s in dunking the ball. What vertical height will he attain? The acceleration of gravity is 9.8 m/s². Answer in units of m.

2. Relevant equations

I'm not sure what to use, but I think [itex]\Delta[/itex]x = v_it + 1/2 at² could work.

3. The attempt at a solution

I divided the hang time (0.843s) by two before plugging it in because I just wanted to find the vertical height, but I'm not sure if that's what I should do. I guess initial velocity would be 0?

I arrived at -2.1m. The absolute value of this is incorrect as well.

Solving this way using the given time (not halved) does not yield the correct answer either. I think the problem is initial velocity. What should it be?

I'm really not sure how to do this one. The wording is vague and I feel like there's not enough information. Any help would be really appreciated, thank you guys.

berkeman

The first half of the hang time, the player will be moving up, and the 2nd half of the hang time he will be coming back down.

The only influence on the vertical motion is gravity, which provides a constant acceleration downward. The initial condition is that there is some initial v_0 in the positive y direction (up). You don't need to know the initial velocity or the player's weight to solve this.

What is the equation for v(t) in the y direction given some initial v_0 and the downward acceleration of gravity? And since you know the acceration of gravity = "g", and that the final downward velocity is equal and opposite to the initial upward velocity, you can solve for v_0.

Then given v_0, you can use the equation for y(t) to calculate the maximum height reached by the soles of the player's feet at the top of the jump. Do they give you any info on the height and arm length of the player to get the ball's height at the top of the jump?

Doc Al

Figure it out. You have the time it takes to reach the highest point, which is all you need.

AVReidy

Why isn't the initial velocity zero? Are we to assume the guy flies through the floor and up to the highest point?

Doc Al

If the initial velocity were zero, he'd just stand there. He's got to jump, which gives him a non-zero velocity as his feet leave the floor.

AVReidy

I must have solved for initial velocity with the wrong equation. I plugged it in to my main equation and it was wrong both when I used 1/2 time and given time.

I used v_f = v_i + at

This is the best equation I know for this job, so if it's the right one I must be wrong again by assuming final velocity is 0.

berkeman

The final velocity at the *top* of the jump (at half of the "hang time") is indeed zero. The final velocity at the end of the hang time is not zero.

Doc Al

Assuming the final velocity is 0 is perfectly fine (when you use 1/2 the total time) and should give you the right answer for Vi. Show exactly what you plugged in.

AVReidy

My math background consists of Algebra I and Geometry, so this isn't obvious to me. I think I need to use a system of equations to solve for initial velocity, though...

Edit: I used the total time by accident. That helps a lot. Now I need to plug initial velocity into the main equation and solve using half the time, I guess?

Doc Al

Nah... that simple equation from your last post is all you need.

AVReidy

I keep getting a negative number (-1.33) when I plug in initial velocity. I really don't understand this problem, nothing seems to work.

Doc Al

Show exactly what you did. How could you plug in initial velocity--that's the unknown that you're solving for.

AVReidy

I solved for initial velocity using that simple equation.

0 = v_i + (-9.8)(0.4215)
4.1307 = v_i

Then I plugged that in.

Doc Al

Looks good to me.

Show the details.

AVReidy

I forgot to square the second time. Thank you very much, you have been a great help. This one really stumped me