3. A ball is thrown up vertically, with a velocity of 20m/s. Calculate the maximum height
reached and the time taken to reach it. Calculate also the time taken for the ball to reach
the ground again.

Explanation:

u=20m/s

v=0m/s

g= 9.8m/s^2

h=?

v^2 = u^2 -2gh

0^2 = 20^2 -2 ×9.8×h

19.6h=400

19.6h/19.6 = 400/19.6

h=20.41m

solving for time taken for the ball to reach the ground again T =the time to reach maximum height + time the Ball takes to fall to the ground .

The time t to reach maximum height = The time t to fall to the ground

T = t+t

T=2t

u=20m/s

v=0m/s

g= 9.8m/s^2

t=?

v= u-gt

0=20-9.8t

9.8t=20

9.8t/9.8 = 20/9.8

t= 2.04seconds

T=2t

T=2(2.04)

T=4.08 seconds is the time the ball takes to reach the ground again.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.