A ball is thrown upward with an initial velocity of 48 ft/sec from a height 640ft. It's height h, in feet, after t seconds is given by h (t) = - 16t^2 + 48t+640. After how long will the ball reach the ground?

The ball will reach the ground in ___seconds

Question

Maleah Carrillo

Physics

14 November, 05:50

A ball is thrown upward with an initial velocity of 48 ft/sec from a height 640ft. It's height h, in feet, after t seconds is given by h (t) = - 16t^2 + 48t+640. After how long will the ball reach the ground?

The ball will reach the ground in ___seconds

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Answers (1)

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Javion · Answer 1

We equate h (t) to 0 to find the time at which the ball reaches the ground.

16t² - 48t - 640 = 0

t² - 3t - 40 = 0

t = 8, t = - 5

Time cannot be negative so second value neglected.

The ball will reach the ground in 8 seconds.

A ball is thrown upward with an initial velocity of 48 ft/sec from a height 640ft. It's height h, in feet, after t seconds is given by h (t) = - 16t^2 + 48t+640. After how long will the ball reach the ground?The ball will reach the ground in ___seconds

A ball is thrown upward with an initial velocity of 48 ft/sec from a height 640ft. It's height h, in feet, after t seconds is given by h (t) = - 16t^2 + 48t+640. After how long will the ball reach the ground?

The ball will reach the ground in ___seconds