Question

An object is projected with a velocity of $20m/s$ making an angle of $45$ with horizontal. The equation for the trajectory is $h=Ax-B{x}^2$, where $h$ is height, $x$ is horizontal distance, $A$ and $B$ are constants. The ratio $A:B$ is- $($$g=10m/s^2$$)$

• A. $1:5$
• B. $5:1$
• C. $1:40$
• D. $40:1$

Answer 1

Answer: (D)

$40:1$

horizontal and vertical component of velocity are same and equal to $20cos{45}^o=10\sqrt{2}$

hence horizontal distance is $x=10\sqrt{2}t$

and vertical distance is$s=ut+\frac{1}{2}a{t}^2=10\sqrt{2}t-\frac{1}{2}g{t}^2$

therefore $h=10\sqrt{2}t-5t^2=A\times 10\sqrt{2}t-B\times 200t^2$

$10\sqrt{2}A=10\sqrt{2}$

$A=1$

$200B=5$

$B=\frac{1}{40}$

therefore $A:B=\frac{1}{\frac{1}{40}}=40:1$

Alternatively, $y=x(A-Bx)=0$ when particle reaches the ground.

Hence, $R=\frac{A}{B}$

But, $R=\frac{u^2\times \sin \left(2\theta \right)}{g}=\frac{{20}^2\sin {90}^0}{10}=40$

$\Rightarrow \frac{A}{B}=40:1$