Solving a Physics Problem: Car Drives Off a Cliff

[eliassiguenza]

Homework Statement

A stunt man drives a car at a speed 20m/s off a 30m high cliff. The road leading to the cliff is inclined upward at an angle of 20 degrees.

How far from the base of the cliff does the car land?

What is the car's impact speed?

The Attempt at a Solution

I thought of just doing some old fashion R = V initial ^2 sin 2 alpha/g

since V = dx/dt

dt = dx/V

=30/ Sin alpha

now that I know dt y multiply by v cos alpha to then add it to R.
Apparently I'm wrong, could someone help me?

 

[jhae2.718]

The Attempt at a Solution

I thought of just doing some old fashion R = V initial ^2 sin 2 alpha/g

You are correct; you can use $$r=\frac{v_0^2\sin(2\theta)}{g}$$.

To find the impact velocity, what kinematic equation relates range and time?

Then, what equation gives velocity in y as a function of time?

 

[eliassiguenza]

[/QUOTE]
now that I know dt y multiply by v cos alpha to then add it to R.[/QUOTE]

Sorry I was meant to type and I multiply now, at impact shouldn't be
Vfinal = V initial + 1/2 gt ^2
?


Thanx For Helping! =)

 

[jhae2.718]

now that I know dt y multiply by v cos alpha to then add it to R.

Sorry I was meant to type and I multiply now, at impact shouldn't be
Vfinal = V initial + 1/2 gt ^2
?Thanx For Helping! =)

$$\frac{1}{2}gt^2$$ will give you distance. The equation for velocity as a function of time is $$v(t)=v_0+gt$$, where the sign of g depends on what direction you call positive.