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?
should be written __

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.

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation:

30 people weigh 160 pounds or more