# Projectile motion and trajectory

1. Oct 9, 2011

### Alvl tay

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

A car is drove off a cliff at 8 m/s. it lands in the sea 3 seconds later.

2. Relevant equations

a) how far does it land?
b)how high is the cliff? (g = 10m/s^2)
c) at what angle relative to the horizontal does it land?

3. The attempt at a solution
convention chosen : upwards is positive
initial velocity= 0m/s
a= -g
Vx=8m/s

a) FROM HORIZONTAL COMPONENT

Vx= x(displacement)/ t( time)
8m/s = x/3s
x=24m

b) FROM VERTICAL COMPONENT
using the eq'n s=ut + 0.5at^2
ie. y = 0.5 gt^2

y = 0.5 (-10)(9)
y= - 45 m

c) ** my problem is that im not sure which eq'n to use

since its relative to the horizontal there is only one... i think.

Vx = Vt cosθ
i have t= 3s
i have x=24m
i dont have V ,or at least i dont know what V exactly is.

Last edited: Oct 9, 2011
2. Oct 9, 2011

### ehild

It will be clear if you make a drawing indicating the initial and final positions of the car. You need the angle of the displacement vector, not the angle of the initial velocity which is horizontal.

ehild

3. Oct 27, 2011

The x-component of the velocity is just the initial 8m/s.

The final velocity of a uniformly accelerated object is
Vf=Vi+gt=0+(-9.8m/s^2)*3s=-29.4m/s
which is the y-component.

Then the angle should be the angle made up from those vector components: tan(8/24), or is it tan(24/8), can't think straight about it at the moment.

4. Oct 28, 2011

### Alvl tay

i solved it already :/ a long time ago.
am i supposed to delete the post then?

5. Oct 28, 2011

### ehild

You should have written that the problem was solved.

ehild

Last edited: Oct 28, 2011