Projectile motion and gravitational acceleration problem

[imhereyeah]

During a physics class on the planet Xoltac, young Greels are asked to determine the gravitational acceleration by using a particle launcher that has a muzzle velocity of 10 m/s. They find that the maximum horizontal range of the particles is 20m. What is the gravitational acceleration on Xoltac? part two: If a Greel were to aim his particle launcher at 60 degrees above the horizontal, what would the range be?

I don't see how to find the gravitational acceleration unless the height off the ground from which the particle is shot is known, which it isn't.

 

[HallsofIvy]

You know, of course, that x= v0 cos(theta) t and
y= v0 sin(theta)t- (g/2) t².

In this problem we only know that v0= 10 m/s- theta is unknown.
In any case, the projectile reaches its "range" when y= 0 again: that means v0 sin(theta)t- (g/2) t²= 0. The solutions are, of course, t= 0 and t= (v0 sin(theta))/((g/2)= (2 vo sin(theta))/g.
At that time x= vo cos(theta)(2 vo sin(theta))/g=
(2v0²sin(theta)cos(theta)/g: the range.

We know that when v0= 10, the "maximum horizontal range" is 20 m.
Since the only variable is theta, differentiate the above with respect to theta to determine the value of theta that gives the maximum range. When you put that into the equation, the only "unknown" left is g so you can solve for that.

Once you know g, you can put that value, together with theta= 60 degrees in the above formula to find the range.

 

[imhereyeah]

I see what I was doing wrong, thank you.