# Calculate the speed with which the ball hits the ground

1. Feb 6, 2006

### discombobulated

I need some help with this mechanics question please,

A ball is thrown vertically upwards with speed 10ms/s from a point 2m above horizontal ground.
a) calculate the length of time for which the ball is 3m or more above the ground.
b) calculate the speed with which the ball hits the ground.

so..i think i have to use the equation s=ut + 1/2 at2
u= 10
a= -9.8
but i'm confused about s, should it be 2m or 3m?

2. Feb 6, 2006

### VietDao29

discombobulated, this is an Introductory Physics kind of problem, not a Precalculus Mathematics one. :)
That's neither 2m nor 3m.
s=ut + 1/2 at2
This equation will return the displacement the object has travelled.
The u is the initial velocity of that object, t is the time the objevt has spent travelling, a is its acceleration.
Now, if the ball goes from 2m above the ground to 3m above the ground. What's its displacement? What's s?
Having s, u, a, you'll be able to find t, right?
From there, do you know how long does the ball is 3m or more above the ground?
Can you go from here?
Can you do part b?

3. Feb 6, 2006

### discombobulated

Thanks, so s=1 then I worked out t = 1.82 by using the quadratic equation to solve and finding the difference.
I think i need to use v2 =u2+2as for part b, but which value of s should i use?
I tried 2m but then i got v=6.26 whichis wong as the correct answer is 11.8.

Last edited: Feb 6, 2006
4. Feb 6, 2006

### VietDao29

The answer you got for a looks good. But don't forget to add the measure unit to t, it should read: t = 1.82 s, not just t = 1.82.
No s is not 2. Let's think about it. You choose a = g = -9.8 m / s2, which means the positive y is upward.
So from 2m above the ground, the ball flies up, reaches its highest point, then falls back down and finally reaches the ground. So from 2m above the ground, the ball finally reaches the ground, what's the displacement, is it positive 2m? Or what is it?
Can you go from here? :)