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If an object is propelled upward from a height of 72 feet at an initial velocity of 90 feet per second, then its height h after t seconds is given by the equation h = − 16 t 2 + 90 t + 72 . After how many seconds does the object hit the ground?

User Ravi Anand
by
8.1k points

1 Answer

3 votes

Answer:

6.34 seconds.

Explanation:

The object will hit the ground when h = 0.

-16t^2 + 90t + 72 = 0

8t^2 - 45t - 36 = 0

We can then use the quadratic formula to solve.

[please ignore the A-hat; that is a bug]


(45±√(45^2 - 4 * 8 * -36) )/(2 * 8)

=
(45±√(2025 + 1152) )/(16)

=
(45±√(3177) )/(16)

=
(45±56.36488268)/(16)

(45 - 56.36488268) / 16 = -0.7103051678

(45 + 56.36488268) / 16 = 6.335305168

Since the time cannot be negative, the object will hit the ground after about 6.34 seconds.

Hope this helps!

User Qingfeng
by
8.3k points

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