Question

If an object is related from top of 200m high building find its velocity and time when it hits the ground.(g=(10m/s^2)

Answer 1

The velocity at the ground is 63.25 m/s and time taken is 6.325 s.

Step-by-step explanation:

Given:

As the object is released, the initial velocity is,
`u=0\ ms^(-1)`

Displacement of the object is,
`d=200\ m`

To find:

Velocity at the bottom,
`v=?`

Time to reach the bottom,
`t=?`

The acceleration of the object is due to gravity and hence equal to
`a=g=10\ m/s^2`

Now, using the following equation of motion:

`v^2=u^2+2ad\\v^2=0+2(10)(200)\\v^2=4000\\\textrm{Taking square root both sides}\\v=√(4000)=63.25\ m/s`

Now, using the equation of motion relating time and velocity:

`v=u+at\\63.25=0+10t\\t=(63.25)/(10)=6.325\ s`

Therefore, the velocity at the ground is 63.25 m/s and time taken is 6.325 s.