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40 votes
40 votes
A NASA test rocket is launched from the top of a 101 foot cliff with an initial

velocity of 116 feet per second. The function h(t) = -16t2 +116t + 101 gives the
height, h(t), in feet of the rocket t seconds after it's launched.
How long does it take for the rocket to reach the ground?


User Gavin Gilmour
by
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1 Answer

12 votes
12 votes

Answer:

4.018secs

Explanation:

Given the height reached by the test rocket by the expression

h(t) = -16t^2 +116t + 101 gives the

The rocket will reach the ground when h(t) = 0

Substitute h(t) = 0 into the expression

0 = -16t^2 +116t + 101

Multiply through by -1

16t^2 -116t - 101 = 0

Factorize using the general formula

t = 116±√(116)²-4(16)(-101)/4(16)

t = 116±√(13,456+6,464)/64

t = 116±√(19,920)/64

t = 116±141.138/64

t = 116+141.138/64

t = 257.138/64

t = 4.018secs

Hence it will take the rocket 4.018secs to reach the ground

User Nathan Cox
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