Question

18. A ball is thrown vertically upwards with an initial velocity of 49 n/s. calculate the maximum height
attained by the ball and time take by it before it reaches the ground. Take g * 9.8 m/s​

Answer 1

Answer:

Initial velocity (u) = 49 m/s    Final velocity (v) = 0 m/s

Gravity in toward down = + 9.8 m/s2  Gravity in toward up = - 9.8 m/s2

(i) We know, V2 - u2 = 2gs

              or,  (o)2- (49)2 = 2 X 9.8 X S

              or,  s = - (49)2/ 2 X 9.8 = 122.5 m

Maximum height = 122.5 m

(ii) We know, v = u + gt

              or   0 = 49 +(- 9.8) X t

              or   9.8 X t =  49

              or, t = 49/ 9.8 = 5 s

If, time for upward direction = time for downward direction

Then, total time taken by the ball to return back = 5 + 5 = 10 s

Answer 2

Answer:

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