Question

A particle is thrown vertically up with speed of 40 m/s from the top of a building of height 100 m . The magnitude of displacement of particle during t=0 to t=9 s is

Answer 1

Answer:

The displacement of the particle is 0 from the ground.

Explanation:

Given:

• Speed of the particle upward direction, u=40 m/s
• Height of then building, 100 m

Now Using equation of motion under gravity in upward direction taking direction top be positive y direction we have

$y=ut-\dfrac{gt^2}{2}$

The total taken by the particle when falls back down on the ground will be the time of flight of the particle after this particle will be the ground.

$0=40\times t-\dfrac{9.8t^2}{2}\\\\t=8.16\ \rm s$

So the particle will be on the ground after this time with zero displacement .As the given time interval is 0 to 9 s so the displacement of the particle will be zero from the ground.