Question

A ball is dropped from the top of a building 122.5m. Find the time taken to reach the ground and the velocity with which it strikes the ground. [ Take g=9.8m/s2 ]

Answer 1

Answer:

Initial velocity, u = 0 m/s

Acceleration due to gravity, g=9.8 m/s

2

Time taken to reach the ground, t = 2.5 sec

Height, h = ?

Using relation,

s=u t+

2

1

gt

2

s=0×2.5+

2

1

×9.8×2.5×2.5

s=0+4.9×2.5×2.5

s = 30.625 m

Explanation:

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Answer 2

Explanation:

Given Initial velocity of ball, u=49 m/s

Let the maximum height reached and time taken to reach that height be H and t respectively.

Assumption: g=9.8 m/s2 holds true (maximum height reached is small compared to the radius of earth)

Velocity of the ball at maximum height is zero, v=0

v2−u2=2aH                          

0−(49)2=2×(−9.8)×H         

⟹H=122.5 m               

v=u+at

0=49−9.8t            

⟹t=5 s

∴ Total time taken by ball to return to the surface, T=2t=10 s