Amotorcycle stunt driver zooms off the end of a cliff at a speed of 30 meters per second. if he lands after 0.75 seconds, what is the height of the cliff?
a. what do you know?
b. what do you need to solve for?
c. what equation(s) will you use?
d. what is the solution to this problem?

A. speed = 30 m/s, time taken to land = 0.75 seconds, g = 9.81 m/s^2
b. we need to solve for the height of the cliff, that is the vertical distance the motorcycle traveled from when it went off the cliff till it landed.
c. newton's equation of motion is used, s = vi*t + 1/2*a*t^2. 
d. the vertical component of the initial velocity of the motorcycle is 0. the acceleration is 9.81m/s^2 and the time taken t is 0.75 s. subtituting these in the equation from c. we have s = (0.5)*(9.81)*(0.75)^2 = 2.8 m

The answer is 12, for the points