Physics · Motion in Plane

JEE Main 2025 — 22 January, Morning Shift — Question 65

A particle is projected at an angle of $30^{\circ}$ from horizontal at a speed of

$60 \mathrm{~m} / \mathrm{s}$ . 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

$\qquad$ .[0pt] [Take, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

Answer: 5

Numerical answer — enter this value.

{} $\mathrm{S}_{1}=30 \times 1-\frac{1}{2} \times 10 \times 1=25$

$\mathrm{S}_{3}=30+\left(\frac{-10}{2}\right) \times(2 \times 3-1)=5$

$\frac{\mathrm{S}_{1}}{\mathrm{~S}_{3}}=\frac{25}{5}=5$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.