Physics Problem?

Question

A basketball player is throwing the ball in the air. The Basketball is thrown directly upward from the ground and returns to its starting postition in 0.80 secs.

a) How long did the ball take to reach its maximum hieght?

b) What was the initial velocity of the ball?

c) How high did the ball go?

Answer 1

a. Total air time is .8 seconds.. that means it reaches apex at .4 seconds.
b. 9.8(.4) = Vo, Vo = 3.92 m/s
c. Yf = Yi + Voy(t) + 1/2(a)(t^2) Yf = 0 + 3.92(.4) - 4.9(.4)^2
I get .784 meters for max height but seems low, I dunno I checked my math seems right.

Answer 2

The time to go up and to come down to its starting position is equal. So, if the total time is 0.8 secs, the time taken to reach maximum height = 0.4 secs.

If a ball starts from rest, it achieves a velocity v in time t given by v = u + at

v = 0 + g x 0.4 = 0.4g = 3.92 m/sec assuming g = 9.8 m/sec^2

When the ball reached its starting position after 0.8 secs, its velocity will be exactly same as it had when going up. So, starting velocity of the ball = 3.92 m/sec

c) v^2 - u^2 = 2as is the formula to be used.

0 - 3.92^2 = 2 x 9.8 x s

s = 3.92 x 3.92 /19.6 = 0.784 m

Answer 3

a) The total time is .8s. It takes the basketball .4s to reach its maximun hieght and another .4s to return to its starting point


b) Vf = at + Vi

Vf = final velocity
t = time
a = acceleration
Vi = initial velocity

when the ball reaches its maximun height, its final velocity is 0m/s and it takes .4s to get there.

0 = -9.8(.4) + Vi
0 = -3.92
Vi = 3.92m/s


c) Xf = 1/2at^2 + Vi*t + Xi

Xf = final position
Xi = initial position

Xf = 1/2 (-9.8)(.4)^2 + (3.92)(.4) + 0
Xf = 0.784m