Q123 / 160 · AP EAMCET 2024 (19 May Shift 2) · Physics

A2 kg ball thrown vertically upward and another 3 kg ball projected with certain angle $(\theta \neq 90)$ both will have same time of flight. Then the ratio of their maximum heights is

  • A
    $2: 3$
  • B
    $3: 2$
  • C
    $\sqrt{3}: 2$
  • Correct answer
    D
    $1: 1$

Solution

For vertical upward projection, $\mathrm{T}_1=\frac{2 \mathrm{u}_1}{\mathrm{~g}}$ For projectile motion, $\mathrm{T}_2=\frac{2 \mathrm{u}_2 \sin \theta}{\mathrm{~g}}$ $\begin{aligned} & \mathrm{T}_1=\mathrm{T}_2 \Rightarrow \frac{2 \mathrm{u}_1}{\mathrm{~g}}=\frac{2 \mathrm{u}_2 \sin \theta}{\mathrm{~g}} \Rightarrow \mathrm{u}_1=\mathrm{u}_2 \sin \theta \\ & \therefore \frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\mathrm{u}_1^2 / 2 \mathrm{~g}}{\frac{\mathrm{u}_2^2 \sin ^2 \theta}{2 \mathrm{~g}}}=\left(\frac{\mathrm{u}_1}{\mathrm{u}_2 \sin \theta}\right)^2=1: 1 \end{aligned}$

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

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