A body is thrown from the surface of the earth with velocity 'V' $\frac{\mathrm{m}}{\mathrm{s}}$. The…

A body is thrown from the surface of the earth with velocity 'V' $\frac{\mathrm{m}}{\mathrm{s}}$. The maximum height above the earth's surface upto which it will reach is $(\mathrm{R}=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity)
  1. $\frac{\mathrm{VR}^{2}}{\boldsymbol{g R}-\mathbf{V}}$
  2. $\frac{\mathrm{V}^{2} \mathrm{R}}{2 \mathrm{gR}-\mathrm{V}^{2}}$
  3. $\frac{2 \sigma R}{V^{2}(R-1)}$
  4. $\frac{\mathrm{VR}}{2 \sigma R-V}$

Solution

$\begin{aligned} \frac{1}{2} \mathrm{mu}^{2} &=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \\ &=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \mathrm{V}^{2} &=2 \mathrm{GM}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \mathrm{V}^{2} &=2 \mathrm{gR}^{2}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right] \\ \mathrm{V}^{2} &=2 \mathrm{gR}\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \quad \frac{\mathrm{V}^{2}}{2 \mathrm{gR}} &=\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \quad \therefore \quad \frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}=\frac{2 \mathrm{gR}}{\mathrm{V}^{2}} \\ \frac{\mathrm{R}}{\mathrm{h}}+1 &=\frac{2 \mathrm{gR}}{\mathrm{V}^{2}} \\ \therefore \frac{\mathrm{R}}{\mathrm{h}} &=\frac{2 \mathrm{gR}}{\mathrm{V}^{2}}-1=\frac{2 \mathrm{gR}-\mathrm{V}^{2}}{\mathrm{V}^{2}} \therefore \mathrm{h}=\frac{\mathrm{Ru}^{2}}{2 \mathrm{gR}-\mathrm{V}^{2}} \end{aligned}$ .

Asked in: MHT CET 2020 (16 Oct Shift 1)

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