John R.

asked • 07/07/22

Physics Help Please

A ball is thrown upward with an initial velocity of 48 feet per second from an initial height of 64 feet. The acceleration due to gravity is a(t) = −32 feet per second squared.


(a) Find the velocity function, v(t), of the ball at time t.


(b) Find the position function, s(t), of the height of the ball at time t.


(c) At what time does the ball reach its maximum height?


(d) What is the maximum height reached by the ball?

1 Expert Answer

By:

John R.

I'm quite confused about that first equation
Report

07/07/22

Paul M.

tutor
The first equation is the "standard" equation of motion of bodies near the surface of the earth. s(t) is the distance traveled at is positive for distance away from the earth in this case, v is velocity with v0 the initial velocity. g is a physical constant called the acceleration due to gravity and the minus sign indicates motion toward the earth. I hope that helps lessen your confusion.
Report

07/08/22

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