JEE Mains · Physics · STD 11 - 3.2 motion in plane
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, \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A 5
- B 6
- C 7
- D 8
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{S}_1=30 \times 1-\frac{1}{2} \times 10 \times 1=25 \\ & \mathrm{~S}_3=30+\left(\frac{-10}{2}\right) \times(2 \times 3-1)=5 \\ & \frac{\mathrm{~S}_1}{\mathrm{~S}_3}=\frac{25}{5}=5\end{aligned}\)
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