Question

Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following and select proper option $(g=10{\text{m/s}}^2)$:
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \text{Angle of projection}& \text{(p)}& 20\text{m}\\ \text{(B)}& \text{Angle of velocity}& \text{(q)}& 80\text{m}\\ & \text{with horizontal}& & \\ & \text{after}4s& & \\ \text{(C)}& \text{Maximum height}& \text{(r)}& {45}^{\circ }\\ \text{(D)}& \text{Horizontal range}& \text{(s)}& {\tan }^{-1}\left(\frac{1}{2}\right)\end{array}$

• A. $A\to s,B\to s,C\to q,D\to p$
• B. $A\to r,B\to r,C\to q,D\to p$
• C. $A\to r,B\to r,C\to p,D\to q$
• D. $A\to s,B\to r,C\to p,D\to q$

Answer 1

The correct option is C $A\to r,B\to r,C\to p,D\to q$

Given
$y=x(1-\frac{x}{80})$
Comparing with the standard form $y=x\tan \theta (1-\frac{x}{R})$,
$\Rightarrow \tan \theta =1\Rightarrow \theta ={45}^{\circ }$
$R=80\text{m}$
We have,
$4H=R\tan \theta$
$4H=80\left(1\right)$
$\Rightarrow H=20\text{m}\Rightarrow \frac{u_y^2}{2g}=20\Rightarrow u_y=20$
Since, $\tan \theta =1\Rightarrow u_x=u_y=20$
At $t$ seconds,
$\tan \alpha =\frac{u_y-gt}{u_x}\Rightarrow \tan \alpha =\frac{20-40}{20}$ at $t=4s$
$\Rightarrow \alpha ={\tan }^{-1}(-1)={45}^{\circ }$ below the horizontal.