A projectile projected at a speed 'u' making theta with horizontal is at the same height 'h' from ground at time instants time =t1 and time=t2.Then h is
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Uy=0 in vertical direction, acceleration in vertical =g, then h = Uyt +1/2at²
t is from t1 to t2 then t =t2-t1 .
so, final answer will be 1/2g(t2-t1)².
t is from t1 to t2 then t =t2-t1 .
so, final answer will be 1/2g(t2-t1)².
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