A particle is projected at an angle of 30° from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h₀ and height traversed in the last second, before it reaches the maximum height, is h₁. The ratio h₀: h₁ is __________. [Take, g = 10 m/s² ] Numerical answer type. Work it out, then open the answer.

Answer: 5. Vertical component uy = 60sin 30° = 30 m/s; time to top = 30/10 = 3 s. First second: h₀ = 30(1) - 1/2(10)(1)² = 25 m Last second before the top (reverse motion from rest): h₁ = 1/2(10)(1)² = 5 m h₀: h₁ = 25: 5 = 5

A particle is projected at an angle of 30° from horizontal at a speed… | JEE Main 2025 (22 Jan, Shift 1)

Q9JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Time of Flight, Maximum Height & Range

A particle is projected at an angle of $30^\circ$ from horizontal at a speed of $60\text{ m/s}$ . The height traversed by the particle in the first second is $h_0$ and height traversed in the last second, before it reaches the maximum height, is $h_1$ . The ratio $h_0 : h_1$ is __________.

[Take, $g = 10\text{ m/s}^{2}$ ]

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

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Answer: 5

Vertical component $u_y = 60\sin 30^\circ = 30\text{ m/s}$ ; time to top $= \frac{30}{10} = 3\text{ s}$ .
First second: $h_0 = 30(1) - \frac{1}{2}(10)(1)^2 = 25\text{ m}$
Last second before the top (reverse motion from rest): $h_1 = \frac{1}{2}(10)(1)^2 = 5\text{ m}$
$h_0 : h_1 = 25 : 5 = 5$

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