I guess 0 is Theta which I will call T
Vi = U sin T (initial vertical speed up)
u = U cos T (horizontal speed)
v = 0 at top
v = U sin T - g t
so at top t = U sin T / g
h = Vi t - (1/2) g t^2
at top h = U sin T t - (1/2)g t^2
= U sin T (U sin T / g) - (1/2) g (U sin T / g)^2
= U^2 sin^2 T / g - (1/2)U^2 sin^2 T / g
= (1/2)U^2 sin^2 T / g
a lead shot is projected from the ground level with a velocity U at an angle 0 to the horizontal. Given the time t for the lead shot to reach its maximum height as t =U Sin0/g where 'g' is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h max = U²Sin²/2g?
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