JEE Advance - Physics (2021 - Paper 1 Online - No. 6)
A projectile is thrown from a point O on the ground at an angle 45$$^\circ$$ from the vertical and with a speed 5$$\sqrt 2 $$ m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g = 10 m/s2.
The value of x is _______________.
The value of x is _______________.
Answer
7.5
Explanation
Range, $$R = {u_x} \times T = {{2{u_x}{u_y}} \over g} = {{2 \times 5 \times 5} \over {10}} = 5$$ m
Time of flight, $$T = {{2{u_y}} \over g} = {{2 \times 5} \over {10}} = 1$$ s
Both particle have no vertical velocity after splitting so both will take same time to reach the ground.
$$\because$$ Time of motion of one part falling vertically downwards is 0.5 s
$$\Rightarrow$$ Time of motion of another part, $$t = 0.5$$ s
From momentum conservation, pi = pf
2m $$\times$$ 5 = m $$\times$$ v
v = 10 m/s
Displacement of other part in 0.5 s in horizontal direction,
$$ = v\left( {{T \over 2}} \right) = 10 \times 0.5 = 5$$ m = R
$$\therefore$$ Total distance of second part from point O is $$x = {{3R} \over 2} = 3 \times {5 \over 2}$$
$$ \Rightarrow $$ x = 7.5 m
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