Q101 / 150 · MHT CET 2024 (03 May Shift 1) · Physics

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 $\left(\sin 90^{\circ}=1, \sin 60^{\circ}=\cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\cos 60^{\circ}=\frac{1}{2}\right)$

  • Correct answer
    A
    $4: 3$
  • B
    $3: 4$
  • C
    $4: 1$
  • D
    $3: 2$

Solution

Maximum height attained by ball $\mathrm{A}, \mathrm{h}_1=\frac{\mathrm{u}^2}{2 \mathrm{~g}}$ Maximum height attained by ball B, $\begin{aligned} \mathrm{h}_2 & =\frac{\mathrm{u}^2 \sin ^2\left(60^{\circ}\right)}{2 \mathrm{~g}} \cdot \ldots\left(: \theta=90^{\circ}-30^{\circ}=60^{\circ}\right) \\ & =\frac{3 \mathrm{u}^2}{8 \mathrm{~g}} \\ \therefore \quad \frac{\mathrm{~h}_1}{\mathrm{~h}_2} & =\frac{\mathrm{u}^2}{2 \mathrm{~g}} \times \frac{8 \mathrm{~g}}{3 \mathrm{u}^2}=\frac{8}{6}=\frac{4}{3} \end{aligned}$

Hence, the correct answer is option: 'A'.

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