JEE Main 2025 Jan 22 Shift 1, Physics Q25: Motion In Two And Three Dimension

JEE Main2025Jan 22, Shift 1Physics

A particle is projected at an angle of $3 0^{\circ}$ from horizontal at a speed of 60 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 m/s $^{2}$ ]

Your answer

Solution

Initial vertical component: $u_{y} = 60 sin 3 0^{\circ} = 30$ m/s. Time to peak: $T = u_{y} / g = 3$ s.

Height in the first second (from $t = 0$ to $t = 1$ ):

$h_{0} = u_{y} \cdot 1 - \frac{1}{2} g (1)^{2} = 30 - 5 = 25$ m.

Height in the last second before peak (from $t = 2$ to $t = 3$ ): use $S_{n} = u_{y} - \frac{g}{2} (2 n - 1)$ with $n = 3$ :

$h_{1} = 30 - \frac{10}{2} (2 \cdot 3 - 1) = 30 - 25 = 5$ m.

$∴ \frac{h _{0}}{h _{1}} = \frac{25}{5} = 5$ .

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JEE Main 2025 Jan 22 Shift 1, Physics Q25: Motion In Two And Three Dimension

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