Question

a man is standing on top of a building 100m 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 1st and 2nd ball is +15m at t=2sec.the gap is found to remain constant.calculate velocity with which balls were thrown and exact time interval between their throw

Answer 1

Final Answer : 20 m/s , 10m/s


Most Important : Ball is thrown in upward direction.

#Steps and Understanding :
1) Let the initial velocity of first ball be "u" m/s .

And,
The only possibility for the gap to remain constant is that :
Both balls has same velocity after second ball is thrown.

2) Implies speed of first ball when second ball is thrown is (u) /2 .

First Ball :
Distance covered ,S = 15 m
Initial velocity, u = u m/s (say)
Final velocity, v = u/2 m/s

By Newtons Third equation of motion,
${v}^{2} = {u}^{2} + 2( - g)s \\ = > {( \frac{u}{2}) }^{2} = {u}^{2} - 2gs \\ = > \frac{3}{4} {u}^{2} = 2gs \\ = > {u}^{2} = \frac{4}{3} \times 2\times 10 \times 15 \\ = > {u}^{2} = 400 \: \\ = > u = 20 \: m \div s$
3) Therefore, Velocity of first ball : 20 m/s in upward direction.
Velocity of second ball is 10m/s in upward direction.

Answer 2

Answer:

Explanation:

Let the first ball be thrown with the velocity and second ball with a time interval . Taking vertical motion of ball first for time , we have

                                      ...()

Taking upward motion of second ball for time , we have

                                      ...()

∴                    

                             

As per the question, ; and   [as ]

Then,                

or                    

or                    

The velocity of the other ball = = = 10 m/s. Time interval = 1 s