MHT CET · Physics · Motion In Two Dimensions
A ball 'A' is projected vertically upwards with certain initial speed. Another ball ' \(B\) ' of same mass is projected at an angle of \(30^{\circ}\) with vertical with the same initial speed. At the highest point, the ratio of potential energy of ball \(A\) to that of ball B will be
\((\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}\) \(=\frac{1}{2})\)
- A \(4: 3\)
- B \(3: 4\)
- C \(4: 1\)
- D \(3: 2\)
Answer & Solution
Correct Answer
(A) \(4: 3\)
Step-by-step Solution
Detailed explanation
Maximum height attained by ball \(\mathrm{A}, \mathrm{h}_1=\frac{\mathrm{u}^2}{2 \mathrm{~g}}\)
Maximum height attained by ball B,
\(\begin{aligned}
\mathrm{h}_2 & =\frac{\mathrm{u}^2 \sin ^2\left(60^{\circ}\right)}{2 \mathrm{~g}} \cdot \ldots\left(: \theta=90^{\circ}-30^{\circ}=60^{\circ}\right) \\
& =\frac{3 \mathrm{u}^2}{8 \mathrm{~g}} \\
\therefore \quad \frac{\mathrm{~h}_1}{\mathrm{~h}_2} & =\frac{\mathrm{u}^2}{2 \mathrm{~g}} \times \frac{8 \mathrm{~g}}{3 \mathrm{u}^2}=\frac{8}{6}=\frac{4}{3}
\end{aligned}\)
Maximum height attained by ball B,
\(\begin{aligned}
\mathrm{h}_2 & =\frac{\mathrm{u}^2 \sin ^2\left(60^{\circ}\right)}{2 \mathrm{~g}} \cdot \ldots\left(: \theta=90^{\circ}-30^{\circ}=60^{\circ}\right) \\
& =\frac{3 \mathrm{u}^2}{8 \mathrm{~g}} \\
\therefore \quad \frac{\mathrm{~h}_1}{\mathrm{~h}_2} & =\frac{\mathrm{u}^2}{2 \mathrm{~g}} \times \frac{8 \mathrm{~g}}{3 \mathrm{u}^2}=\frac{8}{6}=\frac{4}{3}
\end{aligned}\)
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