Question

A ball is thrown vertically upwards with a velocity of 49m/s calculate the maximum height to which it rises and the total time taken to return to the surface of the earth​

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 a 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

Explanation: