JEE Advanced 2021 · Paper 1 · Physics Q5

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PhysicsNumerical+2Question 5 of 57

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Question Stem for Question Nos. 5 and 6

A projectile is thrown from a point O on the ground at an angle $45^\circ$ from the vertical and with a speed $5\sqrt2$ 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 the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g = 10\ \text{m/s}^2$ .

The value of $t$ is ____.

Your answer

Answer and worked solution

Official answer: 0.45

Top of the trajectory. The launch is $45^\circ$ from the vertical at $5\sqrt2$ m/s, so $u_x=u_y=5$ m/s. The highest point is reached after 0.5 s, at height $H=\frac{25}{20}=1.25$ m, with $x=2.5$ m.

The split. Falling 1.25 m from rest takes $\sqrt{2(1.25)/10}=0.5$ s. So the part that falls vertically in 0.5 s has zero velocity just after the split.

Momentum conservation for two equal halves then gives the other part a horizontal velocity of $2u_x=10$ m/s and no vertical velocity. It also falls 1.25 m from rest, so
$t=0.5\text{ s}.$

Topics:Projectile motion Conservation of linear momentum

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JEE Advanced 2021 · Paper 1, Physics question 5