# I know im thick

1. Aug 22, 2006

### russs598

I know im thick!!!

I know this sound stupid but my kid has brought his homework home and although im OK at higher maths i seem to have lost the ability to do the basics!!!!
Can anyone help with this.......

A car leaves a quay and ends in the water 6m away from the foot of the quay. The quay is 5m above water level.

Q. What is the time taken for the car to hit the water? Hence determine the velocity with which it hits the quay.

Cheers

2. Aug 22, 2006

### neurocomp2003

is there a time component to where teh car splashed?
Its projectile motion. That you need to look up.

x=v_x0*t; x=6;
y=the acceleration formula with a=g....mmm can't remmeber but i think its
y+v_y0*t+g/2*t^2; y=5;

solve for v_x0, v_y0...and then do the pythagorean to get the speed. or if they want the velocity vector then your done.

3. Sep 12, 2006

### mtanti

The car falls vertically at the acceleration of gravity so the horizontal speed and distance of the car won't matter.

Assuming g=10m/s^2,
s = 5m, u = 0m/s, a = 10m/s^2, t = ?

s=ut + 0.5a(t^2)
t = sqrt (2s/a)
= 1s

therefore using v=s/t
v = 5m / 1s = 5m/s

Just keep in mind that projectiles have 2 seperate speeds, the horizontal and the verticle and they don't mix up unless you want to find the resultant force.