First slide

Rectilinear Motion

Question

A stone is thrown vertically upward. When the stone is at point A, its distance from a certain point O is $6 \sqrt{5}$ (OA=6 $\sqrt{5}$ ) and the component of velocity along OA is nonzero. When it is at point B(OB= 10 m), the velocity at B is zero. When it is at point C(OC=6 m), the component of velocity of the particle along OC is zero. If the velocity of projection of the stone is v₀ = 5 n ms^-1, then the value of n is __________

Moderate

Solution

From figure,
$\begin{array}{l} A C = \sqrt{(6 \sqrt{5})^{2} - 6^{2}} = 6 \sqrt{5 - 1} = 12 m \\ and C B = \sqrt{10^{2} - 6^{2}} = 8 m \end{array}$
$∴$ The maximum height attained by the ball is
H=AB=12 m+8 m = 20 m
$\begin{array}{l} ∵ v^{2} = u^{2} - 2 g H \\ Or 0^{2} = v_{0}^{2} - 2 g H \\ ∴ v_{0} = \sqrt{2 g H} = \sqrt{2 \times 10 \times 20} = 20 {ms}^{- 1} = 5 n \\ ∴ n = 4 \end{array}$

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