A particle is projected at an angle of $30^\circ$ from horizontal at a speed of $60\ \text{m/s}$. 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 $g = 10\ \text{m/s}^2$]
Numerical value type. Enter your answer.
Answer: 5
Vertical component $u_y = 60\sin30^\circ = 30\ \text{m/s}$; time to reach the top $= 30/10 = 3\ \text{s}$.
First second: $h_0 = 30(1) - \tfrac{1}{2}(10)(1)^2 = 25\ \text{m}$.
The last second before the top is the reverse of a free fall of $1\ \text{s}$ from rest: $h_1 = \tfrac{1}{2}(10)(1)^2 = 5\ \text{m}$.
$h_0 : h_1 = 25 : 5 = 5 : 1$, so the answer is $5$.
Solution by Sreeraj P, M.Sc Physics