$\begin{array}{1 1}(A)\;2hu \\(B)\;\frac{2hu^2}{gb^2} \\(C)\;\frac{1}{2}hu\\(D)\;0 \end{array} $

Horizontal distance covered by ball =nb

Vertical distance covered by ball =nh

If the time taken to hit the edge of nth step is t, then

$nb= ut$

$nh= \large\frac{1}{2}$$gt^2$

Eliminating t, we get,

$nh = \large\frac{1}{2}$$g \bigg(\large\frac{nb}{u}\bigg)^2$

$n= \large\frac{2hu^2}{gb^2}$

Hence B is the correct answer.

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