How Do You Solve Projectile Motion for a Car Driving Off a Cliff?

[Alvl tay]

Homework Statement

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

Homework 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?

The Attempt at a Solution

convention chosen : upwards is positive
initial velocity= 0m/s
a= -g
V_x=8m/s

a) FROM HORIZONTAL COMPONENT

V_x= 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 I am not sure which eq'n to use

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

V_x = Vt cosθ
i have t= 3s
i have x=24m
i don't have V ,or at least i don't know what V exactly is.

Help please?

 

[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

 

[Head_Unit]

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.

 

[Alvl tay]

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

 

[ehild]

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

You should have written that the problem was solved.

ehild