Question

The turd is launched at a speed of 53
m/s at an angle of 28 degrees above
the horizontal. How far horizontally
from the catapult should a portal be
placed in order to intercept the turd
before it hits the ground?

Answer 1

This question can be solved by using the formulae of the projectile motion.

The portal should be placed at a horizontal distance of "237.4 m" from the catapult.

The horizontal range formula of the projectile motion can be used here to find the horizontal distance:

$R = \frac{v^2Sin\ 2\theta}{g}$

where,

R = Horizontal range = ?

v = launch speed = 53 m/s

θ = launch angle = 28°

Therefore,

$R = \frac{(53\ m/s)^2Sin(2*28^o)}{9.81\ m/s^2}$

R = 237.4 m

Learn more about projectile motion here:

https://brainly.com/question/11049671?referrer=searchResults

The attached picture shows the projectile motion.