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 $\mathrm{h}_{0}$ and height traversed in the last second, before it reaches the maximum height, is $h_{1}$. The ratio $h_{0}: h_{1}$ is ____. [Take, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ ]

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Kinematics

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}

]

Answer: 5

The vertical component of the launch velocity is

u_{y} = 60 sin 30° = 30 m/s

, with

g = 10 m/s^{2}

.

Height covered in the first second:

h_{0} = u_{y} t - \frac{1}{2} g t^{2} = 30 (1) - \frac{1}{2} (10) (1)^{2} = 30 - 5 = 25 m .

Time to reach maximum height is

t_{ma x} = \frac{u _{y}}{g} = 3 s

, so the last second before maximum height is the interval from

t = 2 s

t = 3 s

. Using

h (t) = u_{y} t - \frac{1}{2} g t^{2}

h (2) = 60 - 20 = 40 m, h (3) = 90 - 45 = 45 m,

h_{1} = h (3) - h (2) = 45 - 40 = 5 m .

The ratio is

h_{0} : h_{1} = 25 : 5 = 5.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Integer answer