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.

 

[rwisz]

Solving this physics problem involves using the equations of motion to determine the distance and impact speed of the car as it drives off the cliff. To start, we can use the equation d = v0t + (1/2)at^2 to find the distance the car travels in the horizontal direction before reaching the edge of the cliff. In this case, v0 is the initial velocity of the car, which is 20 m/s, and a is the acceleration due to gravity, which is -9.8 m/s^2. We also know that the angle of the road is 20 degrees, so we can use trigonometry to find the vertical component of the car's initial velocity, which is v0y = v0sin(20). This will be the initial velocity in the vertical direction, and we can use it in the equation vf = v0 + at to find the time it takes for the car to reach the edge of the cliff.

Once we have the time, we can use it to find the horizontal distance traveled using the first equation. Then, we can use the equation vf^2 = v0^2 + 2ad to find the impact speed of the car when it reaches the ground. In this equation, vf is the final velocity, which we can find using the initial velocity in the vertical direction and the time we calculated earlier.

Using these equations, we can solve for the distance from the base of the cliff and the impact speed of the car. It is important to always double check your units and make sure they are consistent throughout the calculations. Additionally, it is helpful to draw a diagram and label all known quantities before starting the calculations. This will help to ensure that you are using the correct equations and plugging in the correct values.