Question

A projectile is projected at an angle of ${45}^{\circ }$ to the horizontal. The slope of trajectory of the body varies with time $t$ as

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Trajectory of projectile, $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$
$\frac{dy}{dx}=\text{Slope}=\tan \theta -\left(\frac{2g}{2u^2{\cos }^2\theta }\right)x$
$(\because x=ucos\theta t)$

$\frac{dy}{dx}=\text{Slope}=\tan \theta -\left(\frac{2g}{2u{\cos }^{}\theta }\right)t$
$\Rightarrow Y=c-mt$

As $t$ increases, slope is positive and decreases, becomes zero at the highest point and then it becomes negative in the downward journey.