Answer and Explanation:
As per the question:
When the stone is thrown from the cliff top and hits the ground below eventually:
R = [tex]v_{o}\sqrt{\frac{2H}{g}}[/tex]
where
[tex]v_{o}[/tex] = initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
Now,
(a) Displacement of the stone is given by the horizontal range:
R = [tex]v_{o}\sqrt{\frac{2H}{g}}[/tex]
where
[tex]v_{o}[/tex] = initial velocity
H = height
g = acceleration due to gravity
R = horizontal Range
(b) Speed just prior to the impact is given by the third equation of motion:
[tex]v = \sqrt{v_{o}^{2} + 2gH}[/tex]
where
v = final velocity
(c) Time of flight is given by the second eqn of motion where the initial velocity is considered to be 0 then:
[tex]H = v_{o}T + \frac{1}{2}gT^{2}[/tex]
[tex]H = 0.T + \frac{1}{2}gT^{2}[/tex]
T = [tex]\sqrt{\frac{2H}{g}[/tex]
