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Question 4 (1 point)

A ball is thrown from a mountain 960 feet high, into the air with an upward velocity of 64 ft/s. When did the ball hit the

ground? And What was the max height it reached?

User Chroman
by
6.3k points

1 Answer

7 votes

Answer:

The ball will hit the ground at 10 seconds and its maximum height is 1024 feet.

Explanation:

Height of a mountain, h = 90 feet

Upward velocity of a ball, u = 64 ft/s

Its height as a function of time can be written by using the second equation of motion as follows :


f(x) = -16t^2 + 64t + 960 .....(1)

When it will hit the ground, f(x) = 0


-16t^2 + 64t + 960=0\\\\16(-t^2+4t+60)=0\\\\-t^2+4t+60=0

It is quadratic equation.

t = -6 s and t = 10 s

Neglecting -6 seconds, the ball will hit the ground at 10 seconds.

For maximum height,

Put f'(x) = 0


-32t+64=0\\\\\\t=(64)/(32)\\\\t=2\ s

Put t = 2 in equation (1)


f(2) = -16(2)^2 + 64(2) + 960\\\\=1024\ \text{feet}

Hence, the ball will hit the ground at 10 seconds and its maximum height is 1024 feet.

User Tadas Davidsonas
by
7.1k points
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