The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre) of a projectile is $y=3 x-0.8 x^2$. The time of flight of the projectile is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ ).

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Aicharya · Accepted Answer

For a projectile, $y=3 x-0.8 x^2$....(i) $\begin{aligned} & \text { At } x=R, y=0 \\ & \therefore \quad 0=3 R-0.8 R^2 \\ & \Rightarrow \quad R(0.8 R-3)=0 \\ & \therefore \quad R=\frac{3}{0.8}=\frac{15}{4} m \end{aligned}$
Also, $\mathrm{y}=\mathrm{x} \tan \theta-\frac{\mathrm{gn}^2}{2 \mathrm{u}_{\mathrm{x}}^2}$...(ii) Comparing eq(i) with eqn (ii), we get $\frac{\mathrm{g}}{2 \mathrm{u}_{\mathrm{x}}^2}=0.8 \Rightarrow \mathrm{u}_{\mathrm{x}}=\sqrt{\frac{10}{2 \times 0.8}}=\frac{5}{2} \mathrm{~m} / \mathrm{s}$
Now, $R=u_x \cdot T \Rightarrow T=\frac{R}{u_x}=\frac{\frac{15}{4}}{\frac{5}{2}}=1.5 \mathrm{~s}$

The relation between the horizontal displacement $x$ (in metre) and the vertical displacement $y$ (in metre)…

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