20.8k views
5 votes
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 __

2 Answers

2 votes

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

Explanation:

User Fruggiero
by
8.0k points
6 votes

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.

User Divyani Yadav
by
7.5k points

No related questions found