A particle projected from level ground is at a height of $15$ m above the ground at $t = 1$ s and again at $t = 3$ s. If its horizontal range is $60$ m, its speed of projection is ______ m/s. (Take $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 25
Concept. The motion is symmetric about the highest point, so the particle is at equal heights at times equally spaced about $\dfrac T2$. Hence $T = t_1 + t_2$.
Step 1. Time of flight. $T = 1 + 3 = 4$ s.
Step 2. Vertical component. $T = \dfrac{2u_y}{g} \Rightarrow u_y = 20$ m/s. (Check: $y(1) = 20 - 5 = 15$ m.)
Step 3. Horizontal component. $u_x = \dfrac{R}{T} = \dfrac{60}{4} = 15$ m/s.
Step 4. Speed. $u = \sqrt{20^2 + 15^2} = 25$ m/s.
Answer. $25$.
Solution by Sreeraj P, M.Sc Physics