Question

A ball is thrown upward from a height of 432 feet above the​ ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. ​a) Find​ s(t), the function giving the height of the ball at time t. ​b) How long will the ball take to reach the​ ground? ​c) How high will the ball​ go?

Answer 1

Answer;

A)S(t)=96t-16t² +432

B)it will take 9 seconds for the ball to reach the ground.

C)864feet

Explanation:

We were given an initial height of 432 feet.

And v(t)= 96-32t

A) we are to Find​ s(t), the function giving the height of the ball at time t

The position, or heigth, is the integrative of the velocity. So

S(t)= ∫(96-32)dt

S(t)=96t-16t² +K

S(t)=96t-16t² +432

In which the constant of integration K is the initial height, so K= 432

b) we need to know how long will the ball take to reach the​ ground

This is t when S(t)= 0

S(t)=96t-16t² +432

-16t² +96t +432=0

This is quadratic equation, if you solve using factorization method we have

t= -3 or t= 9

Therefore, , t is the instant of time and it must be a positive value.

So it will take 9 seconds for the ball to reach the ground.

C)V=s/t

Velocity= distance/ time

=96=s/9sec

S=96×9

=864feet