Question

An object is projected with a velocity of $20m{s}^{-1}$ making an angle ${45}^o$ with the horizontal. The equation of the trajectory is $y=Ax+B{x}^2$, where $y$ is the height and $x$ is horizontal distance. Then $\left|\frac{A^2}{4B}\right|$ is $(g=10m{s}^{-2})$

• A. $10m$
• B. $5m$
• C. $40m$
• D. $1m$

Answer 1

Answer: (A)

$10m$

$y=20\sin 45t-5t^2$ -(1)
$x=20\cos {45}^{o}t$
$t=\frac{x}{20\cos {45}^o}$ -(2)

putting value of $t$ from (2) in (1) , we get,

$y=x-\frac{5x^2}{{20}^2\cos {45}^2}=x-x^2\times \frac{5}{400\times \frac{1}{2}}$
$\Rightarrow x-\frac{x^2}{40}=y$

$\Rightarrow y=Ax+B{x}^2$
$\Rightarrow A=1,B=-\frac{1}{40}$
So, $\left|\frac{A^2}{4B}\right|=10m$