Question

A dirt bike lanches off of a ramp that is 8 feet tall. The inital velocity of the dirt bike is 20 feet per second the hight is given by the equation h=16t^2+20t+8. What is the maximum hight the dirt bike reaches?

Answer 1

The correct answer is 8 feet.

Explanation:

Height of the dirt bike is given by the equation h = 16
`t^(2)` + 20 t +8.

Initial velocity of the dirt bike is 20 feet per second

Initial height of the dirt bike when it launches off the ramp is 8 feet.

To find the maximum h, we differentiate the function h with respect to t and equate the result to zero.

⇒
`(d)/(dt)`h = 0 = 32t + 20

and second order differentiation of h gives
`(d^(2)h)/(dt^(2))` = 32.

This gives the value of the function maximum at t = -0.625 seconds.

This is absurd as the value of t cannot be negative.

Thus the dirt bike is going to fall off the ramp and is not going to gain any height, making the height at t = 0 the maximum height attained by it.