Question

A heavy particle is projected with a velocity at an angle with the horizontal into a uniform gravitational field. The slope of the trajectory of the particle varies not according to which of the following curve?

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (C), (D)

The correct options are
B
C
D

Equation of the trajectory for the projectile,
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
Slope of the trajectory,
$\frac{dy}{dx}=tan\theta -\frac{2gx}{2u^{2}co{s}^2\theta }$
The above equation can be compared with the straight line equation, $y=mx+c$
where $c=tan\theta$ and $m=-\frac{2g}{2u^{2}co{s}^2\theta }$
Since both options B and D have positive slope for variation with x. Hence the slope of trajectory does not vary according to the variation shown by option B and D
Vertical component of velocity using 1st equation of motion, $\begin{array}{}{V}_y=usin\theta -gt& \text{(2)}\end{array}$
Horizontal component of velocity, $\begin{array}{}{V}_x=ucos\theta & \text{(3)}\end{array}$
Dividing $\left(2\right)$ and $\left(3\right)$
Slope$=\frac{V_y}{V_x}=tan\theta -\frac{gt}{ucos\theta }$
The above equation matches with the straight line equation between slope and time$\left(t\right)$ i.e
$y=-mx+c$
This implies option C also does not show the variation of slope of trajectory since the curve shows either quadratic or cubic variation.
Hence correct options are B,C and D