Question

A ball is thrown vertically up with a velocity of 49m/s. Calculate 2M (i)The maximum height reached by the stone
(ii) The position of the stone from the ground after 7 seconds it is thrown?

Answer 1

Answer:

https://www.toppr.com/ask/en-in/question/a-ball-is-thrown-vertically-upwards-with-a-velocity-of/

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