Question

A tennis ball is thrown straight up with an initial speed of 22.5 m/s. It is caught at the maximum height of its arc.
a. How high does the ball rise? 25.83 m
b. How long does the ball remain in the air?

Answer 1

At the maximum height of the arc, the speed of the ball is 0 m/s (i.e. the final speed). You already know the initial speed and the acceleration, so plug those variables into a=Δv/Δt and you have your answer!

Answer 2

@matt101

I'm not getting the time though

Answer 3

0 = 22.5^2 -19.6d
d= 25.82

Answer 4

Distance = initial velocity * time - 4.9t^2

Answer 5

25.82 = 22.5t - 4.9t^2

Answer 6

t = 2.25 and 2.33

Answer 7

The answer key says twice that. Is this because I used 0 as the final velocity (maxima) and the maxima is the midpoint between the roots?

Answer 8

which t do i use

Answer 9

Use the equation I gave you. Try to avoid using the quadratic to solve for time in these sorts of questions because you get two answers.

Answer 10

If the answer is twice that, the question isn't clear. The question says the ball is caught at when it reaches it's maximum height and that's it. If the ball wasn't caught and instead was allowed to return to its original height (the height from which it was thrown), then yes, you would double the time to account for the up and the down.

Answer 11

A = Delta V/ Delta T
9.8 = 22.5/t
9.8t = 22.5
22.5/9.8 = t
t = 2.295

Answer 12

Yup looks good!

Answer 13

Thank you.