Answer to Question #169282 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=-16t²+64t. v₀=64ft/sec. g=32ft/sec.

istantaneous velocity of thrown object is v=v₀-gt.

a) v(1)=64-32*1=32ft/sec

b) v(3)=64-3*32=-32ft/sec, - sign means the ball is falling.

c) in order to find the time which the height of the ball will be the highest, we need to find the time which the velocity of the ball will be 0.

v(t)=64-32*t=0. t=2.

d) after 2 seconds the ball will have the highest point. inserting 2 second to s equation gives us the highest point. s=-16*2²+64*2=64 feet.

e) from a) v(1)=32ft/sec. from b) v(3)=-32ft/sec.

f) time taken to rise is equal to that of falling. rising time is 2sec, falling time is also 2 sec. it means it takes 4 sec the ball reaches the ground.

g) final velocity of thrown object is always equals to the thrown initial velocity, in our case it is 64 ft/sec.