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A ball 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?
Should be written __

Answer :

Answer:

Substitute 0 for s (the distance from the ground) to find the time it takes for the ball to hit the ground.

substitute 29 for time for the second one

Step-by-step explanation:

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 :

[tex]s=64t-16t^2[/tex]

t is the time taken

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

[tex]64t-16t^2=0[/tex]            

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,

[tex]64t-16t^2>28[/tex]    

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

[tex]0.5<t<3.5[/tex]

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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