A body is projected from the ground at an angle of 45 ° with the horizontal. Its velocity after 2s is 20 ms ⁻¹. The maximum height reached by the body during its motion is __________ m. (use g = 10 ms ⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 20. At 45°: ux = uy = u/√(2). After 2 s: vy = u/√(2) - 20. v² = u²/2 + (u/√(2) - 20)² = 400 u² - 20√(2) u = 0 ⇒ u = 20√(2) m/s H = (u²sin ² 45°)/(2g) = (800 × 1/2)/(20) = 20 m So the maximum height is 20 m.

A body is projected from the ground at an angle of 45 ° with the… | JEE Main 2022 (24 Jun, Shift 2)

Q13JEE Main 2022 · 24 JUN · SHIFT 2Medium

Velocity, Momentum & Energy in Flight

A body is projected from the ground at an angle of 45 $^\circ$ with the horizontal. Its velocity after 2s is 20 ms ${}^{-1}$ . The maximum height reached by the body during its motion is __________ m. (use g = 10 ms ${}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 20

At $45^\circ$ : $u_x = u_y = \frac{u}{\sqrt{2}}$ . After 2 s: $v_y = \frac{u}{\sqrt{2}} - 20$ .
$v^2 = \frac{u^2}{2} + \left(\frac{u}{\sqrt{2}} - 20\right)^2 = 400$
$u^2 - 20\sqrt{2}\,u = 0 \Rightarrow u = 20\sqrt{2}\text{ m/s}$
$H = \frac{u^2\sin^2 45^\circ}{2g} = \frac{800 \times \frac{1}{2}}{20} = 20\text{ m}$
So the maximum height is 20 m.

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Velocity, Momentum & Energy in Flight question from JEE Main 2022 (24 Jun, Shift 2)

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