Thus, the initial speed of the stone is 76.246m/s
Height of the cliff = 208m
Distance travelled during last half second = 48.5 m
According to the second equation of motion:
[tex]s1= vt + 1/2 gt^{2}[/tex]
[tex]v= (2x_{1} - gt^{2})/2t[/tex]
v= [tex]\frac{2(48.5) - (9.8)(0.5)^{2} }{2(0.5)}[/tex]
v= 94.55 m/s
if the initial speed of the ball is u,
then again from the equation of motion, we get initial speed as
u=[tex]\sqrt{v^2 - 2gh}[/tex]
=[tex]\sqrt{94.55^2 -2(9.8)(208-48.5)}[/tex]
u= 76.246 m/s
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