Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Answers (1)

best_answer

Let the distance travelled by the stone which is dropped from the top upto the instant when the two stones meet be x 

Initial velocity u = 0

Acceleration a = g = 9.8 m s-2

Using the second equation of motion

\s=ut+frac12at^2\ x=0 imes t+frac12 imes 9.8 imes t^2\ x=4.9t^2(i)

The distance travelled by the stone which is projected vertically upwards from the ground up to the instant when the two stones meet would be equal to 100 - x

Initial velocity = 25 m s-1

Acceleration a = -g = -9.8 m s-2

Using the second equation of motion

\s=ut+frac12at^2\ 100-x=25 imes t+frac12 imes (-9.8) imes t^2\ 100-x=25t-4.9t^2\ x=4.9t^2-25t+100(ii)

Equating x from (i) and (ii) we get

4.9t2 = 4.9t2  - 25t + 100

25t = 100

t = 4s

x = 4.9t2

x = 4.9 X 42

x = 78.4 m

100 - x = 100 - 78.4 = 21.6 m

The stones meet after a time of 4 seconds at a height of 21.6 metres from the ground.

Posted by

Deependra Verma

View full answer

Similar Questions

Latest Question