Question

An object is projected with velocity $20m{s}^{-1}$ making an angle of ${45}^0$ with horizontal. The equation for its trajectory is $h=Ax-{Bx}^2$, where h is the height and x the horizontal distance. The ratio of constants A:B is?

• A. 1:5
• B. 5:1
• C. 1:40
• D. 40:1

Answer 1

The correct option is D 40:1

When a particle is projected at an angle $\theta$ with the horizontal with velocity u,the trajectory is given by:

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$y=x\tan {45}^{\circ }-\frac{10x^2}{2\times 400{\cos }^{2}45}$

$=x-\frac{10x^2}{800\times 1/2}$

$=x-\frac{10x^2}{400}$

$=x-\frac{1}{40}{x}^2$

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

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