# Homework Help: Need Help with Projectile Motion Problem

1. Sep 27, 2007

### Xiongster

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

I have a problem here and I just can't seem to figure it out. The problem is, A ball is thrown horizontally from a height of 24.28 m and hits the ground with a speed that is 4.0 times its initial speed. What was the initial speed?

2. The attempt at a solution

I've found the velocity in the y-direction by using the equation

v^2 = vo^2 + 2 a x, the velocity is 21.81 m/s.

I've found the time it takes for the ball to drop using the equation

v = vo + a t, t comes out to be 2.22s.

But here on out, I don't know what to do next, i know that vx is constant and acceleration is zero in the x-direction, but I'm unable to get anything more. Any help will be appreciated, thanks in advance!

2. Sep 27, 2007

### l46kok

lol I'm seeing a lot of projectile problems posted today.

You've done the major part of calculation. The y-direction of velocity occurs when the projectile lands, meaning at that particular point there is no x-direction of velocity. And initially, when the ball is thrown you only have x direction of velocity. Does this give you a clue of how to solve the last part? You only have to do one simple arithmetic.

3. Sep 27, 2007

### l46kok

OOPS OOPS, damn, forget my last response, that was misleading (Damn, I had too many beers last night). You DO have both x and y components of velocities when the ball lands. You've calculated the y-component of velocity and x-component of velocity stays the same, so your total velocity is

$$v = \sqrt{v^2_x + v^2_y}$$

Then the problem says this is 4 times greater than the initial velocity, which only has x-component on it. Set the total velocity equal to $$\frac{v_x}{4}$$, solve for $$v_x$$

Last edited: Sep 27, 2007
4. Sep 27, 2007

### Xiongster

Oh I see, I finally got the answer, thanks a lot!