Question

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Answer 1

Answer:

Given:

The height of the building is =100 m

The first ball is thrown at the instant t=0 and another ball is thrown after a time interval of x.

Let, the initial velocity of first ball=u m/s.

The distance covered by the ball is given by:

S

1

=u(t+x)−

2

1

g(t+x)

2

On the other hand, the second ball is thrown at half the speed as of the first ball. So, the distance covered by the second ball is given by:

S

2

=

2

u

t−

2

1

gt

2

The distance between the two balls at t=2 s is 15 m.

∴S

1

−S

2

=15

Now, S

1

−S

2

=u(t+x−

2

t

)−

2

1

g(t

2

+2xt+x

2

−t

2

)

Substitute the values in above equation:

Consider the time interval x=1 since it is less than 2 sec.

15=u(1+

2

2

)−

2

1

×10(2×1×2+1)

u=

2

40

m/s=20 m/s

Therefore, velocity of first ball=20m/s in upward direction

Velocity of second ball is

2

20

=10m/s in upward direction.

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