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, [tex]u=0\ ms^{-1}[/tex]
Displacement of the object is, [tex]d=200\ m[/tex]
To find:
Velocity at the bottom, [tex]v=?[/tex]
Time to reach the bottom, [tex]t=?[/tex]
The acceleration of the object is due to gravity and hence equal to [tex]a=g=10\ m/s^2[/tex]
Now, using the following equation of motion:
[tex]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[/tex]
Now, using the equation of motion relating time and velocity:
[tex]v=u+at\\63.25=0+10t\\t=\frac{63.25}{10}=6.325\ s[/tex]
Therefore, the velocity at the ground is 63.25 m/s and time taken is 6.325 s.
