Question

7. A ball is dropped from rest. The acceleration due to gravity is 10 m s-2 and the time it takes

for the ball to reach the ground is 5 s. What was the velocity of the ball just before it hit the

ground?​

Answer 1

Explanation:

Initial velocity gained  by ball is u=10m/s upward and the acceleration is downward, 

so first of ball will rise to a height till its velocity becomes zero this height above the *initial* position 

is supposed to be H and H=2gu2=2×10100=5m

then it will come back to initial point then will come back to ground.

So net distance =5+5+20=30m

Option C is correct.

Answer 2

Answer:

Given :

u = 0m/s

g = 10m/s^2

t = 5s

v = ?

Formula :

v = u + gt

solution :

v = 0m/s + 10m/s^2 * 5s

v = 50m/s

hope it helps