# Checking my work

Need Confirmation

## Homework Statement

A ball is thrown upward with an initial speed of 80 ft/sec. How high does it go? What is its speed at the end of 3 sec? How high is it at that time?

## Homework Equations

v=v0+at
x=x0+v0+$$\Delta$$t+.5a$$\Delta$$t2

## The Attempt at a Solution

Note: Only the first equation was taught to me, my teacher isn't the best....

Because it is in ft/sec I convert it and get 24.38 m/s rounded. So..
v=24.38-9.8t [-9.8 m/s because gravity right?]
So, in three seconds the velocity is -5.02 m/s.

So, I am trying to find the max height and that occurs at 0 m/s, so I plug in 0 for final velocity.

0=24.38-9.8t
t=2.49 seconds

Then, I use the second equation:
x=0+[24.38*2.49]+0.5[-9.8*2.492
x= 30.325 meters is the maximum height.

This is new territory for me, so I'd like to make sure I am correct. I have finals in a few days.

Last edited:

Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Homework Helper
Welcome to PF!

Hi Chris-P! Welcome to PF!

That looks ok.

But there are three standard constant acceleration equations to learn.

To find the maximum height (was that part of the question? ) you should use vf2 = vi2 + 2as

Thanks for the welcome :)

The question was a bit unclear as it says "how high does it go" so I assumed maximum height.

Also, thanks for the equation, but when I plugged everything in I got a different answer of 326.5 meters for maximum height :P

tiny-tim