Answer to Question #168965 in Calculus for shein

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

Solution

Let s(t) = -16t²+64t. Then instantaneous velocity at any time t is v(t) = s’(t) = -32t+64.

A.     At t = 1 sec v = 32 ft/sec. v>0 => the ball rising at the end of 1 sec.

B.     At t = 3 sec v = -32 ft/sec. v<0 => the ball falling at the end of 1 sec.

C.     At the highest point v(t) = 0. => -32t+64 = 0 => t = 2 sec.

D.     At t = 2 sec s(t) = 64 ft.

E.      At t = 1 sec v = 32 ft/sec. At t = 3 sec v = -32 ft/sec.

F.      At the ground s(t) = 0. => -16t²+64t = 0 =>  t₁ = 0, t₂ = 4. t₁ = 0 is the start point. At t₂ = 4 ball will return on the ground.

G.     The instantaneous velocity of the ball when it reaches the ground is v(4) = -64 ft/sec. A negative value means that the speed direction is downward.