Question

The equation of projectile is y = px - qx² Then the horizontal range is

• A. p
• B. q
• C. pq
• D. $\frac{p}{q}$

Answer 1

Answer: (D)

$\frac{p}{q}$

Answer 2

The standard equation of the trajectory of a projectile is given by: $y = x \tan \theta \left(1 - \frac{x}{R}\right)$ where $\theta$ is the angle of projection and $R$ is the horizontal range.

The given equation of the projectile is: $y = px - qx^2$

We can rewrite the given equation by factoring out $px$: $y = px \left(1 - \frac{qx}{p}\right)$

Now, comparing this rewritten equation with the standard equation: $px \left(1 - \frac{qx}{p}\right) = x \tan \theta \left(1 - \frac{x}{R}\right)$

By comparing the terms, we can equate:

1. The coefficient of $x$: $p = \tan \theta$
2. The term inside the parenthesis: $\frac{qx}{p} = \frac{x}{R}$

From the second comparison, we can solve for $R$: $\frac{q}{p} = \frac{1}{R}$ $R = \frac{p}{q}$

Thus, the horizontal range of the projectile is $\frac{p}{q}$.