Question

During a physics class on the planet Xoltac, students are asked to determine the gravitational acceleration using a particle launcher that has a muzzle velocity of 10.0 m/s. They find that the maximum horizontal range of the particles is 20.0 m.
What is the gravitational acceleration on Xoltac?
Assume the students now use a launching angle of 60o. What is the maximum height the particles reach?
How does the time to go up to the maximum height compare to the time to come back down?

Answer 1

You should be aware that the maximum horizontal range occurs when a projectile is fired at 45 degrees to the horizontal (assuming we're ignoring air resistance). From v=d/t, we have:

\[vcos \theta={R \over t} \implies t={R \over v \cos \theta}\]
Where vcosθ is the horizontal component of the velocity, R is the range, and t is the time the trip took.

This equation will allow us to solve for t, but we want to find the VERTICAL acceleration. There must be another equation we can use. Let's focus instead on the vertical velocity. You should know that when the projectile reaches its maximum height, the vertical velocity is 0. Moreover, the maximum height occurs halfway through the trip (when the time is t/2). So, from a=v/t, we have:

\[a={v \sin \theta \over {t \over 2}}={2v \sin \theta \over t}\]
You can combine the two equations now to solve for a! What do you get?