Question

At t=0 an object is released from rest at the top of a tall building. At the time to a second object is dropped from the same point. After time t (t > to) from starting the separation between objects become l. There is value of to such that the given separation l is achieved in the earliest possible value of t. Then find the value of to (in seconds) if ! = 45 m, g = 10 m/s2) = ​

Answer 1

Answer:

Solution

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Correct option is

C

11.25 m

After time t, distance covered by first stone is s=

2

1

gt

2

After time (t-1), distance travelled by second stone is s=u(t−1)+

2

1

g(t−1)

2

=20t−20+

2

1

g(t−1)

2

These distance must be equal at the time of overtaking,

Thus,

2

1

gt

2

=20t−20+

2

1

g(t−1)

2

or, 0=20t−20+

2

1Put this t in equation for distance to get s=

2

1

gt

2

=

2

1

10(1.5)

2

=11.25m

g(1−2t)

which gives, 20t−20+5−10t=0 (putting g=10)

or t=1.5

Answer 2

Answer:

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