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.
Answers
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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