Q 11-03-042JEE MainEasy
A ball is projected with a speed of $25$ m/s at an angle of $53°$ above the horizontal. Find its horizontal range in metres. (Take $\sin 53° = 0.8$, $\cos 53° = 0.6$ and $g = 10\ \text{m/s}^2$.)
Numerical value type. Enter your answer.
Answer: 60
Concept. Range $= u_x \times$ time of flight.
Step 1. Components. $u_x = 15$ m/s, $u_y = 20$ m/s.
Step 2. Time of flight. $T = \dfrac{2 \times 20}{10} = 4$ s.
Step 3. Range. $15 \times 4 = 60$ m.
Answer. $60$.
Solution by Sreeraj P, M.Sc Physics