Question

The height y and the distance x along the horizontal plane of a projectile projected on a planet are given by x = 8t and y = 6r - 10r ^ 2 where x and y are in metres and t in seconds. The angle of projection with the horizontal at which projectile is projected is:

Answer 1

$\tan^{-1}\left(\frac{3}{4}\right)$

Answer 2

Given the motion equations:

$x = 8t \quad \text{and} \quad y = 6t - 10t^2,$

we compare with the standard projectile motion formulas:

$x = u\cos\theta \, t, \quad y = u\sin\theta \, t - \tfrac{1}{2}gt^2.$

From $x = 8t$, we have:

$u \cos\theta = 8.$

From $y = 6t - 10t^2$, we get:

$u \sin\theta = 6.$

Thus, the angle of projection $\theta$ is given by:

$\tan\theta = \frac{u\sin\theta}{u\cos\theta} = \frac{6}{8}=\frac{3}{4}.$

Hence, $\theta = \tan^{-1}\left(\frac{3}{4}\right).$