Question

# Two particle held at different height $$a$$ and $$b$$ above the ground are allowed to fall from rest. The ratio of their velocities on reaching the ground is

A
a:b
B
a:b
C
a2:b2
D
b:a

Solution

## The correct option is B $$\sqrt{a}:\sqrt{b}$$Let we assume that both particles takes time $$t_1$$ & respectively to reach the ground.$$S = ut + \dfrac{1}{2} at^2$$$$g = 10 \,m/s$$For first particle, $$a = 5t_1^2$$   ...(1)For second particle, $$b = 5t_2^2$$   ...(2)By dividing $$eq^n (1)/ eq^n (2)$$$$\dfrac{t_1}{t_2} = \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$$Now applying, $$V = u + at$$$$10 t_1 = V-1$$    ...(3)$$10 t_2 = V_2$$   ...(4)Dividing $$eq^n (3)/(4)$$$$\dfrac{V_1}{V_2} = \dfrac{t_1}{t_2} = \dfrac{\sqrt{a}}{\sqrt{b}}$$$$V_1 : V_b = \sqrt{a}:\sqrt{b}$$Physics

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