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QJEEAdv21Phde82: JEE Advanced 2021 Physics - Class 11 JEEAdv21Phde82 Solution: A projectile is thrown from a point O on the ground at an angle …

A projectile is thrown from a point O on the ground at an angle 45 ∘^\circ from the vertical and with a speed 5 2\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 _______________.

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Physics - Laws of Motion Question 10 English Explanation 1
Range, R=ux×T=2uxuyg=2×5×510=5R = {u_x} \times T = {{2{u_x}{u_y}} \over g} = {{2 \times 5 \times 5} \over {10}} = 5 m

Time of flight, T=2uyg=2×510=1T = {{2{u_y}} \over g} = {{2 \times 5} \over {10}} = 1 s

JEE Advanced 2021 Paper 1 Online Physics - Laws of Motion Question 10 English Explanation 2JEE Advanced 2021 Paper 1 Online Physics - Laws of Motion Question 10 English Explanation 3
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=0.5t = 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(T2)=10×0.5=5= v\left( {{T \over 2}} \right) = 10 \times 0.5 = 5 m = R

∴\therefore Total distance of second part from point O is x=3R2=3×52x = {{3R} \over 2} = 3 \times {5 \over 2}

⇒\Rightarrow x = 7.5 m

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