Answer to Question #347310 in Calculus for Stella

Question

A ball thrown straight up is located s feet above the ground at t seconds after it is thrown in accordance with the formula s= 112t-16 t^2. Find a formula for the velocity of the ball and find (a) the time required to reach its highest point, (b) the distance of the highest point above the ground, and (c) the acceleration of the ball at this point.

Answer 1

$s(t)= 112t-16 t^2.$

The velocity can be found as a derivative of position:

$v(t)=\frac{d}{dt}(s(t))=\frac{d}{dt}(112t-16t^2)=112-16\cdot2t=112-32t.$

(a) At the highest point $v=0,$ so we can find the time required to reach the highest point:

$0=112-32t,$

$32t=112,$

$t=\frac{112}{32}=3.5$ s.

(b) The distance of the highest point above the ground:

$s=s(3.5)=112\cdot3.5-16\cdot3.5^2=395-196=196$ f.

(c) The acceleration is a derivative of velocity:

$a(t)=\frac{d}{dt}(v(t))=\frac{d}{dt}(112-32t)=-32$ f/s².

The velocity is constant, so the acceleration of the ball at the highest point is -32 f/s².

Answer: 3.5 s; 196 f; -32 f/s².