JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 25)
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, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
Answer
5
Explanation
_22nd_January_Morning_Shift_en_25_1.png)
Time to reach max. height,
$${v_y} = {u_y} - gt$$
$$O = 30 - 10t \Rightarrow t = 3s$$
We know, $${S_n} = u + {a \over 2}(2n - 1)$$
So, $${h_0} = {S_1} = 30 - {{10} \over 2}\left( {2 \times 1 - 1} \right) = 30 - 5 = 25$$ m
and $${h_1} = S{w_3} = 30 - {{10} \over 2}\left( {2 \times 3 - 1} \right) = 30 - 25 = 5$$
Hence, $${{{h_0}} \over {{h_1}}} = {{25} \over 5} = 5$$
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