# Projectile motion

1. Feb 20, 2006

### brandon26

A particle is projected on a horizontal ground and moves freely under gravity.
The horizontal and vertical components of the initial velocity are 2u and u ms^1 respectively.

The ball lands a distance 80m from the point of projection.

Show that u=14

This is what I did:

Taking motion in the horizontal plane------

2u = 80/t

Taking motion in the vertical plane-----

displacement (s) =0
acceleration (a)= -9.8
initial velocity (u) = u
time of motion (t) = t

therefore using s=ut +(1/2)at^2

I get: 4.9t=u

substituting this into the first equation should give me the correct value of u, but it doesnt. Can someone tell me where i went wrong?

2. Feb 20, 2006

### finchie_88

$$0 = ut - 4.9t^2$$
Assuming that t is non-zero, since that is an obvious solution, $$t = \frac{2u}{g}$$, therefore, since velocity in the horzontial plane is constant, $$4u^2 = 80g$$, then solve that

3. Feb 20, 2006

### HallsofIvy

Staff Emeritus
Why doesn't it? If u= 4.9 t, then t= u/4.9. Putting that into
2u= 80/t, 2u= (80)(4.9)/u so u2= 40(4.9). What do you think that is?