Question

1)

a. A ball is tossed up into the air with an initial speed of 7.7. How long does it take to return to the person's hand?

b. Let's look at the ball in a different way. A ball is tossed up into the air. It takes the ball 9.2 seconds to return to the person's hand. What is the greatest height that the ball reached above the person's hand?

Answer 1

a)

Here we have,

Initial velocity u = 7.7 m/s

Final velocity of the ball, v = 0

Time taken by the ball to reach maximum height will be as,

V = u - gt

t = 7.7/9.8 = (0.78571428571) sec

And here also ,Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the players hands = 2 × 0.7857 = (1.57142857143) sec

Which is the required time to reach persons hand.

b) here time taken to reach maximum height will be as,

t = T/2 = 9.2/2 = 4.6 sec

So that,

Initial velocity here will be as,

U = g×t = 9.8×4.6 =(45.08) m/s

So that maximum height the ball will be reaching is given as,

S = U²/2g = 45.08²/2×9.8 = (103.684) m

Which is our required maximum height above person hand that ball will reach.