Expert: Josh - 6/12/2009

Question
A model rocket is shot into the air and its path is approximated by h = -5t sqaured + 30t where h is the height of the rocket and t is the elasped time in secounds.

1)when will the rocket hit the ground?
2) what is the maximum height of the rocket?


Answer
Hello Ceceli

For 1), you need to find t when the rocket has zero elevation. i.e., you set h=0 in the equation and solve for t.

For 2), the maximum height is found in two steps. Have you learned differentiation yet? If you know how to differentiate any polynomial expression, then you can find the first derivative of h with respect to t. The projectile turns around precisely at the moment when its velocity is zero [motion changes from positive (upward) to negative (downward)]. So, we set dh/dt (viz., the velocity) to 0 and solve for t. Then, we plug in this "t" value into the function h. Here is an illustration.

Step 1: As h=-5t^2+30t, dh/dt=-10t+30. Now we set dh/dt=0. We find -10t+30=0 when t=3.
Step 2: Put t=3 into h=-5t^2+30t. We get 45 (in meters I presume).