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
$u_x = 25 \times 0.6 = 15$ m/s, $u_y = 25 \times 0.8 = 20$ m/s.
Time of flight: $T = \dfrac{2u_y}{g} = 4$ s.
Range: $R = u_xT = 15 \times 4 = 60$ m.
Solution by Sreeraj P, M.Sc Physics