A stunt man drives a car at a speed of 20m/s off a 30 m high cliff. The road leading to the cliff is inclined upward at an angle of 20 degree.
How far from the base of the cliff does the car land?
What is the car's impact speed?

Hi, hsns0098!

If you post what you think might be the answers and the ways that you arrived at them, you'll be more likely to have someone knowledgeable come along to teach you how to arrive at the correct answers, if the answers that you post are incorrect, and also to verify for you that the answers that you've posted are correct, if indeed, they are.

Thanks !

1. Break the speed into horizontal component and vertical component.

2i). Using , find the time of flight of the car.

ii). Use that time and the horizontal component to find the horizontal displacement.

3i). Using , find the vertical component of the car at the bottom of the cliff. (be careful, consider well your signs)

ii). Use Pythagoras' Theorem to recombine the new vertical component and the horizontal component.

I gave you the steps. If you really want help now, post what you did, as Clough said.

I used v^2*sin2theta/g
for the first part but it gave me 26.2 meter. Which is wrong.
can u tell me y it is wrong?
and I couldn't get the second part?
Thanks

1. Well, I never use that formula even if I know it. You see, that formula finds the range of projectiles on the same level as the projected object.

Attachment 25690

R is what you get using your formula, and R is what you are asked to find, which you get using the method I told you.

2. The horizontal velocity remains the same all the time, because there are no horizontal forces acting on the car (neglecting air resistance)

You can find the vertical velocity using the formula I gave you. Since you have the vertical and horizontal final components of the speed, you can find the total velocity, using Pythagoras' Theorem.

Tell me if you got it now, or if I have to say more. :)