Aball is thrown vertically upward with an initial velocity of 64 feet per second. the distance s (in feet) of the ball from the ground after t seconds is s=64t-16t^2.

(a) at what time t will the ball strike the ground?
(b) for what time t is the ball more than 28 feet above the ground?
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Step-by-step explanation:

It is given that,

Initial velocity of the ball, u = 64 ft/s

The distance of the ball from the ground is given by the following relation as :

t is the time taken

(a) Let t is the time when the ball strike the ground. When it strikes the ground, s = 0

On solving the above quadratic equation, we get the value of t as t = 4 seconds. So, at 4 seconds the ball will strike the ground.

(b) Let t is the time when the ball is more than 28 feet above the ground. So,

On solving the above inequalities, we get the value of t as :

So, from 0.5 seconds to 3.5 seconds, the ball is at the height of 28 feet above the ground.

Hence, this is the required solution.

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step-by-step explanation: