Q128 / 160 · AP EAMCET 2024 (22 May Shift 1) · Physics
A boy weighing 50 kg finished long jump at a distance of 8 m . Considering that he moved along a parabolic path and his angle of jump is $45^{\circ}$, his initial $K E$ is
- A960 J
- B1560 J
- C2460 J
- Correct answerD1960 J
Solution
$\mathrm{R}=8 \mathrm{~m}, \theta=45^{\circ}$
$\begin{aligned}
& \therefore \quad R=\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{9.8} \\
& \therefore \quad u^2=9.8 R=9.8 \times 8=78.4
\end{aligned}$
$\therefore \quad$ Initial K.E. of the man,
$(\mathrm{K} . \mathrm{E})_2=\frac{1}{2} \mathrm{mu}^2=\frac{1}{2} \times 50 \times 78.4=1960 \mathrm{~J}$
Hence, the correct answer is option: 'D'.