What equation do I use for this..?

Question

A baseball is hit nearly straight up into the air with a speed of 26 m/s.

How high does it go?

How long is it in the air?

I thought of using v=v0 + at but don't think that's correct.,

Answer 1

Well your right, you do not use that equaiton.

What you should do is use: v^2 = v0^2 - 2gy

Take this equation and solve for y. This will give you an equation that allows you to solve for height. [Note] v = zero when the ball reaches its maximum height. Setting v = 0 in the above equation and then solving for y will give you the max height the ball reaches.

                                y = V0^2 / 2g

This equation is what you need for part (a) of your problem.

There are different ways to find the time it takes for the ball to travel in the air. I will show you, in my opinion, the easiest way to do it.

Find the time it takes for the ball to travel to its maximum height. This will account for half the time. Then to account for the time it is falling back down (the other half) multiply the time you obtained by 2.

Time it takes to travel to max height, or half its distance (v =0):

v = v0 - gt

0 = v0 - gt

t = v0/g   <-- time it takes to travel half the distance

Total time = 2(t)

  

Answer 2

v = ut + 1/2at(^2)

Where, v = 0, u= 26, a= -9.8