Q44 / 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 x is ________.

Correct answer: 7.5

Solution

Figure

So,

 x=R2+r 

Where,R is the range of oblique projectile and r is the range of horizontal projectile.

x=R2+10×2hg

x=u2sin2θg×2+10×0.5

x=5020+5

x=2.5+5

x=7.5 m

Hence, the correct answer is 7.5.

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