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

Answer:

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

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:

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

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

$v=u+at\\63.25=0+10t\\t=\frac{63.25}{10}=6.325\ s$

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