Question

A ball is thrown up with a speed of 19.6ms^-1.
i. What is the acceleration due to gravity acting on it?
ii. Calculate the maximum height it would gain before it begins to fall.
iii. Also, calculate the time it takes to reach this height.
iv. Prove that the time of ascent is equal to the time of descent.​

Answer 1

Answer:

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