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A rocket carrying fireworks is launched from a hill 80 feet above a lake. The rocket will fall into lake after exploding at its maximum height. The rocket's height above the surface of the lake is represented by h = -16t^2 + 64 t + 80. How many seconds after the rocket launched will it hit the lake?

Respuesta :

abidemiokin

Answer:

t = 2secs

Step-by-step explanation:

Given the height above the surface expressed as h = -16t^2 + 64 t + 80.

The velocity of the rocket at its maximum height is zero. Hence;

v(t) = dy/dt = 0

dy/dt = -32t + 64

0 = -32t + 64

32t = 64

t = 64/32

t = 2secs

Hence the rocket will hit the lake after 2 secs

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