Problem 83.

What is the speed (m/sec) needed for a stunt driver to launch from a 20 degree ramp to land 15 m away?

What is his maximum height?

Solution:

Initial speed.

This is the projectile motion. There are two sets of equations, which describe the motion of the projectile (stunt driver):

Set 1: motion along horizontal axis (axis x – see figure). This is the motion with constant velocity. There is only one equation, which describe this motion:

..............................(1)

Here and .

Set 2: motion along vertical axis (axisy – see figure). This is the motion with constant acceleration – free fall motion. There are three equations, which describe this motion. Only two equations are independent, but it is convenient to write all three equations:

................(2)

...........................(3)

...........................(4)

We know that the y-coordinate of the final point (point B) is 0 and the x-coordinate of the final point is 15 m. We substitute these values in equations (1) and (2) and obtain

Then from the first equation we can find :

Substitute this expression into the second equation:

From this equation we can easily find the initial velocity:

Maximum height.

The condition that the projectile is at the point with the maximum height is that the y-component of its velocity at this point is zero. It is easier to find the maximum height from equation (4). Indeed, we substitute in this equation and obtain:

Then

We know , then we can find the maximum height: