Answer to Question #169614 in Calculus for Nics

Question

A ball is thrown vertically upward from the ground with an initial velocity of 64 ft/sec. If the positive direction of the distance from the starting point is up, the equation of the motion is s =-16t²+64t. Let t seconds be the time that has elapsed since the ball was thrown and s feet be the distance of the ball from the starting point at t seconds.

A. Find the instantaneous velocity of the ball at the end of 1 sec. Is the ball rising or talling at the end of 1 sec?

B. Find the instantaneous velocity of the ball at the end of 3 sec. Is the ball rising or falling at the end of 3 sec?

C. How many seconds does it take the ball to reach its highest point?

D. How high will the ball go?

E. Find the speed of the ball at the end of 1 sec and 3 sec.

F. How many seconds does it take the ball to reach the ground?

G. Find the instantaneous velocity of the ball when it reaches the ground.

Answer 1

$s=-16t^2+64t$

A.

$v(t)=s'(t)=-32t+64$

$v(1)=32$

the ball is rising because $v>0$

B.

$v(3)=-32*3+64=-32$

the ball is falling because $v<0$

C.

in highest point: $v=0$ , $-32t+64=0$ , $t=2$

D.

$s(2)=-16*2^2+64*2=64$

E.

the speed of the ball at $t=1$ and at $t=3$ equals $32 feet/sec$ ,

see A. and B.

F.

$s(t)=-16t^2+64t=-16t(t-4)=0$

$t=0$ (moment of throwing)

$t=4$ (moment of falling on the ground)

G.

$v(4)=-32*4+64=-64$