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!