If a rock released with an initial velocity = u
If it falls under gravity 'g' for 't' seconds,
Expression to find the distance covered will be,
[tex]y=ut+\frac{1}{2}t^2[/tex]
If the initial velocity is zero expression for the distance covered will be,
[tex]y=\frac{1}{2}gt^2[/tex]
Here, g = Gravitational pull Or acceleration due to gravity
A). If the distance covered 'y' = 20 m
And t = 4 seconds
By substituting these values in the expression,
[tex]20=\frac{1}{2}g(4)^2[/tex]
[tex]20=8g[/tex]
[tex]g=2.5[/tex] m per sec square
B). Average velocity of a free falling body = [tex]\frac{\text{Initial velocity}+\text{Final velocity}}{2}[/tex]
Initial velocity of the object = 0
Final velocity = [tex]g\times t[/tex]
= [tex]2.5\times 4[/tex]
= [tex]10[/tex] meter per sec.
Therefore, average velocity = [tex]\frac{0+10}{2}[/tex]
= [tex]5[/tex] meter per second
C). Since, the rock hits the bottom after [tex]4[/tex] seconds,
Velocity with which the rock will hit the ground [tex]=gt[/tex]
[tex]=10[/tex] m per sec.
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