Question

A boy throws balls into air at regular intervals of 2 seconds.n the next ball is thrown when the velocity of the first ball is zero. how high do the ball arise above his hand

Answer 1

Answer:

answer of this question is 19.6m

Answer 2

Answer:

The ball rises 19.6 m above his hand

Step-by-step explanation:

Step 1: Velocity of ball is 0 when the ball is at maximum height

Step 2: The interval is 2S, it means that after the 1 st ball reaches max height at that time 2nd ball is thrown

Step 3: Time interval =2 seconds

Step 4: Time ascent = u/g

So, time of ascent= u/g = 2

So u= 19.6 m/s

Step 5: Then maximum height= u2/2g =19.6 x 19.6/2 x 9.8

 

Step 6: Maximum height = 19.6 m