Physics - Kinematics Question with Solution | TestHub

Let the angle of projection be$\text{θ}$



$y=\mathit{ax}-{\mathit{bx}}^2$

To get maxima,$\frac{\text{dy}}{\text{dx}}=0$

$\frac{\text{dy}}{\text{dx}}=\text{a}-\text{b}.\text{2x}=0$

$\Rightarrow \text{x}=\frac{\text{a}}{\text{2b}}$

$y_{\mathit{\text{max}}}=\frac{a^2}{2b}-\frac{{\mathit{\text{ba}}}^2}{4b^2}=\frac{{\text{a}}^2}{\text{4b}}$

Angle of projections can be obtained by slope$\left(\frac{\text{dy}}{\text{dx}}\right)$at$\text{x}=0$

$\Rightarrow$Slope at$\text{x}=0is\text{a}$

$\Rightarrow \text{tan}\theta =\text{a}$

$\Rightarrow \theta ={\text{tan}}^{-1}\left(\text{a}\right)$