Q43 / 57 · JEE Advanced 2021 (Paper 1) · Physics

A projectile is thrown from a point $\mathrm{O}$ on the ground at an angle $45^{\circ}$ from the vertical and with a speed $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, $0.5 \mathrm{~s}$ after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.
The value of t is _________.

Correct answer: 0.5

Solution

Figure

After splitting of projectile in two equal parts one part falls vertically down to the ground. 

So, velocity of another part will be according to conservation of linear momentum,

m×5=m2×0+m2×v

⇒v=10 m s−1

Now time taken by m1, to reach ground, 

since after splitting of projectile m1 will perform horizontal projectile so,

t=2hg

⇒t=2u2sin2θ2g2

⇒t=2×502×10×10×12=0.25

⇒t=0.5 s 

Hence, the correct answer is 0.5.

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