Question

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 1

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

Explanation:

Answer 2

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 :

`s=64t-16t^2`

t is the time taken

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

`64t-16t^2=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,

`64t-16t^2>28`

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

`0.5<t<3.5`

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.