A body thrown vertically upwards from the ground reaches a maximum height ' $H$ '. The ratio of the…

A body thrown vertically upwards from the ground reaches a maximum height ' $H$ '. The ratio of the velocities of the body at heights $\frac{3 \mathrm{H}}{4}$ and $\frac{8 \mathrm{H}}{9}$ from the ground is
  1. $4: 9$
  2. $27: 32$
  3. $3: 2$
  4. $3: 8$

Solution

Maximum height, $\mathrm{H}=\frac{\mathrm{u}^2}{2 \mathrm{~g}} \Rightarrow \mathrm{u}=\sqrt{2 \mathrm{gH}}$
For $\mathrm{h}_1=\frac{3 \mathrm{H}}{4}, \mathrm{v}_1^2=\mathrm{u}^2-2 \mathrm{gh}_1=(\sqrt{2 \mathrm{gH}})^2-2 \mathrm{~g}\left(\frac{3 \mathrm{H}}{4}\right)$ $\therefore \quad \mathrm{v}_1=\sqrt{\frac{1}{2} \mathrm{gH}}$ For $\mathrm{h}_2=\frac{8 \mathrm{H}}{9}, \mathrm{v}_2^2=\mathrm{u}^2-2 \mathrm{gh}_2=(\sqrt{2 \mathrm{gH}})^2-2 \mathrm{~g}\left(\frac{8 \mathrm{H}}{9}\right)$ $\therefore \quad \mathrm{v}_2=\sqrt{\frac{2}{9} \mathrm{gH}}$ $\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\sqrt{\frac{\mathrm{gH}}{2} \times \frac{9}{2 \mathrm{gH}}}=\frac{3}{2}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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