Question

A player throws a ball upwards with an initial speed of 29.4m/s. To what height dose the ball rise and after how long does the ball return to the players hands?​

Answer 1

Answer:

Initial velocity (u) = 29.4 m/s

Acceleration due to gravity (g) = -9.8 m/s^2

Final velocity (v) = 0 m/s

h is Height and t is Time.

So,

v^2 = u^2-2gh

0 = u^2-2gh

2gh = u^2

2*9.8*h = 29.4^2

19.6h = 864.36

h = 864.36/19.6

Height = 44.1 m

Now, when the ball is coming back to ground,

g = 9.8 m/s^2

u = 0 m/s

v = 29.4 m/s

v = u+gt

29.4 = 9.8t

3 seconds = Time

I hope this helps.