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, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]

Solution

$\begin{aligned} & \mathrm{S}_1=30 \times 1-\frac{1}{2} \times 10 \times 1=25 \\ & \mathrm{~S}_3=30+\left(\frac{-10}{2}\right) \times(2 \times 3-1)=5 \\ & \frac{\mathrm{~S}_1}{\mathrm{~S}_3}=\frac{25}{5}=5\end{aligned}$

Hence, the correct answer is 5.