An object projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of 8 s and the object reaches foot after 16 s . The height of the tower is [ $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]

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Aicharya · Accepted Answer

During downwârd journey, $\begin{aligned} & \mathrm{t}_{\mathrm{BC}}=\frac{8}{2}=4 \mathrm{~s} \\ & \mathrm{t}_{\mathrm{AC}}=\frac{16}{2}=8 \mathrm{~s} \end{aligned}$ $\therefore$ Height of the tower is $\mathrm{H}=\frac{1}{2} g\left(\mathrm{t}_{\mathrm{AC}}^2-\mathrm{t}_{\mathrm{BC}}^2\right)=\frac{1}{2} \times 10\left(8^2-4^2\right)=240$

An object projected upwards from the foot of a tower. The object crosses the top of the tower twice with an…

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