Suppose a rocket is ascending at a rate of v(t) = 10 t^ 2 + 14t feet per second

Question

Suppose a rocket is ascending at a rate of v(t) = 10 t^2 + 14t feet per second during the rst 6 seconds of its ight. If the rocket starts on the ground, how long is it before it reaches an altitude of 100 feet? (b) Suppose after 6 seconds the engine shuts down. What is the rockets altitude at this time? Suppose after the engine shuts down, the rocket coasts with a velocity of v(t) = 32(t 6) + v0, where t is the time since launch and v0 is the velocity when the engine shuts down. At what time does it hit the ground, and how fast is it going at that time?

Explanation / Answer

integrating we find that d(t) = 10/3*t^3 + 7t^2

set equal to 100, quadratic equation, t = 2.54s

plug in, 6, 972 ft

can't read the rest of the q

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