Question

Ball A is thrown vertically upward with a speed of 19 .6 m/s from the top of high building . as it passes the edge on the way down , a second Ball B is thrown downward at 19.6 m/s . which of the following option is correct: (1) ball B hits the ground before A . (2) ball A hits the ground before B . (3) the two ball hits the ground at same time. (4) data insufficient

Answer 1

(3) The two balls hits the ground at the same time.
Motion of ball A till it passes the edge of building again :-
u = 19.6 m/s
a = -9.8 m/s
s = 0
v²-u² = 2as
v²-19.6² = 2×-9.8×0
v²-19.6² = 0
19.6² = v²
v² = 19.6²
v = √19.6²
v = ± 19.6
The ball will at s=0 at two time when it starts to move and when it comes back.
When it comes back, velocity should be negative as it is going towards the downwards direction.
Therefore, v = -19.6 m/s at s = 0
Motion of ball A after crossing the edge :-
u = -19.6 m/s
Motion of ball B :-
u = -19.6 m/s as it is also thrown in the downward direction.
Both the balls have same initial velocities at the same time and same amount of gravitational force is acting on both balls.
Therefore, they will take equal time to reach the ground.

Answer 2

The correct answer is option (3) the two balls hit the ground at the same time.

.GIVEN

The velocity of ball A = 19.6 m/s

velocity of the ball B = 19.6 m/s

TO FIND

Which ball hits the ground faster.

SOLUTION

We can simply solve the above problem as flows -

It is given that,

The initial velocity of the Ball A after crossing the edge = -19.6 m/s

when hitting the ground,

The final velocity the of ball A = 0

Acceleration = g = 10m/s

The initial velocity of the ball B = - 19 m/s. (in downward direction)/

Let the height of the building be, h

Since it is given that Ball b is thrown when ball A has reached the edge of the way down and both of them have the same initial velocity.

So, the time taken by both the ball to cover the same height 'h' will also be equal.

Hence, The correct answer is option (3) the two balls hit the ground at the same time.

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