The correct option is A 4 s
20 m
Given:
Initial velocity, u=20 ms−1
Acceleration due to gravity, g=−10 ms−2
Upward direction is taken as positive.
At maximum height, velocity will be zero, (v = 0) as the ball will change the direction of motion. Let the maximum height be H.
Using third equation of motion,
v2=u2+2as
02=202+2×(−10)×H
H=20 m
When the ball returns to the initial point, net displacement is zero. Let the time of flight be T. Using second equation of motion, we get:
s=ut+12at2
0=20×T+12×(−10)×T2
T=0 or T=4 s
Time of flight is more than zero. So, it is 4 s.