Question

calculate the velocity of a body released from 40m heigh when it touches the ground (g=9.8m/s^2)
(1)38m/s
(2)46m/s
(3)28m/s
(4)48m/s​

Answer 1

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/s

2

holds true (maximum height reached is small compared to the radius of earth)

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

v

2

−u

2

=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

Answer 2

Answer:

28 m /s

Explanation:

plz give me points