A particle is projected at an angle of $30^\circ$ 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/s²]

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $u_y=30$ m/s, so the time to the top is $3$ s. $h_0=30(1)-\frac12(10)(1)^2=25$ m. In the last second before the top (motion reversed from rest) $h_1=\frac12(10)(1)^2=5$ m. The ratio is $5:1$, so the value asked is $5$. Answer: 5.

Q. A particle is projected at an angle of $30^\circ$ 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/s²]

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?