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A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is h = -16t2 + 272t + 1344. When will the ball strike the ground?

User Badiya
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2 Answers

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The answer is -16(t+4)(t-21)
User Henry Barker
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2 votes

Answer:

Time taken to reach the ball to the ground is 21 seconds.

Explanation:

Given : A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is
h = -16t^2 + 272t + 1344.

To find : When will the ball strike the ground?

Solution :

The equation for the ball's height 'h' at time 't' is given by,


h(t)= -16t^2 + 272t + 1344

When the ball strike the ground the height of the ball became zero.

Substitute h=0 in the given equation,


-16t^2 + 272t + 1344=0

Taking 16 common,


16(-t^2 + 17t + 84)=0

or
t^2-17t-84=0

Solve by middle term split,


t^2-21+4t-84=0


t(t-21)+4(t-21)=0


(t-21)(t+4)=0


t=21,-4=0

Reject t=-4 as time can never be negative.

Therefore, Time taken to reach the ball to the ground is 21 seconds.

User RyanWilcox
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