Question

Ball A is thrown upward with a velocity of 19.6 m/s. Two seconds later ball B is thrown upward with a velocity of 9.8 m/s. Which ball is first to return to the thrower's hand?

Answer 1

V_a=initial upward velocity of ball A = 19.6 m/s

V_b=initial upward velocity of ball B = 9.8 m/s

time taken by the ball to return to the initial position is given by , ( u is the initial velocity )

time taken by ball A to return to thrower's hand =( 2*19.6) / 9.8

= 4 sec

time taken by ball B to return to thrower's hand =( 2*9.8) / 9.8

= 2 sec

Now ball B is thrown 2 seconds later so, by the time it reaches the thrower's hand,

the total time( 2 + t_b= 4 ) equals to the time taken by the ball A to reach its initial position( i.e thrower's hand).

Both the balls reaches the thrower's hand simultaneously.