Question

A projectile is fired on a horizontal ground. Variation of slope of its trajectory with horizontal
displacement is shown (g = 10 m/s2). Choose the CORRECT statement(s) :​

Answer 1

To show:

Variation of slope of its trajectory of Projectile with horizontal displacement .

Calculation:

The equation of trajectory of Projectile is:

$\boxed{ \sf{y = x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2}( \theta) } }}$

Now , the slope of trajectory will be given by the derivative of y wrt x .

$\therefore \: slope = \dfrac{dy}{dx}$

$= > \dfrac{dy}{dx} = \dfrac{d \bigg \{x \tan( \theta) - \dfrac{g {x}^{2} }{2 {u}^{2} { \cos}^{2}( \theta) } \bigg \}}{dx}$

$= > \dfrac{dy}{dx} = \tan( \theta) - \dfrac{2gx}{2 {u}^{2} { \cos}^{2}( \theta) }$

$= > \dfrac{dy}{dx} = \tan( \theta) - \dfrac{gx}{ {u}^{2} { \cos}^{2}( \theta) }$

Comparing this with a standard equation:

$\boxed{ \sf{y = mx + c}}$

We can say that slope is negative and intercept is positive.