Question

Equation of trajectory of a projectile is given by $y=-x^2+10x$ where $x$ and $y$ are in meters and $x$ is along horizontal and $y$ is vertically upward and particle is projected from origin. Then which of the following options is/are correct?
$(g=10m/s^2)$

• A. Initial speed of particle is $\sqrt{505}m/s$
• B. Horizontal range is $10m$
• C. Maximum height is $25m$
• D. Angle of projection with horizontal is ${\tan }^{-1}\left(5\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Initial speed of particle is $\sqrt{505}m/s$
B Horizontal range is $10m$
C Maximum height is $25m$

$\begin{array}{}y=10x-x^2\text{or}y=10x(1-\frac{x}{10})& \text{(1)}\end{array}$
On comparing equ. $\left(1\right)$ with equation of trajectory,
$y=x\tan \theta (1-\frac{x}{R})$
$\tan \theta =10\Rightarrow \sin \theta =\frac{10}{\sqrt{101}}$
and $\cos \theta =\frac{1}{\sqrt{101}}$
and $R=10m$
$R=\frac{2u^2\sin \theta \cos \theta }{g}\Rightarrow 10=\frac{2u^2\times \frac{10}{\sqrt{101}}\times \frac{1}{\sqrt{101}}}{10}\Rightarrow u^2=505$
$\Rightarrow u=\sqrt{505}m/s$
Maximum height $\left(H\right)=\frac{u^2\sin \theta }{2g}=\frac{505\times 100}{20\times 101}=25m$