A man shoots a gun that is located 500m horizontally from his target, and the target is 100m vertical, what is the angle that the man must shoot from to hit his target? Velocity is 20 m/s

Is this an actual question printed on a test? 20 m/s doesn't match any muzzzel velocity I have heard of.

It was actually a homework question given by my Physics Professor!

Give him an annonomous note saying his questions are goofy. He made something up and doesn't understand guns. The grains of gun powder and weight of the bullet would affect the arc of the shot. This question can not be answered with data given. Love to hear what he thinks the answer should be.

I don't know, could be a spud gun. Stop being picky :p

I don't think it's possible to hit the target at that initial velocity.
Did you state the numbers right?

Yes, the numbers are right, and we had to exclude any type of air resistance. Is taking the inverse tan of the two distances the wrong way to find the angle? That was the only thing I could think of that was close to solving this problem.

No, doing the inverse of tan will give a smaller angle than required.

I've tried it but I meddled up with the numbers... here's it.

Let the required angle be \theta

The horizontal component is 20cos\theta and the vertical component 20sin\theta.

The time is given by :

20cos\theta = \frac{500}{t}

\frac{500}{20cos\theta} = t

Then, you use s = ut+\frac12 at^2

Then I got lost with the equation to find the angle. The displacement is 100, acceleration -9.81, time is just above, u is vertical component.

See below. I plotted the y position when x = 500m against launch angle. It doesn't approach anywhere near 100m.

Hence my earlier question about the wording of the question - it seems unsolvable.

(I got the x axis units on the graph wrong - it's in degrees)

Oh, that's perhaps why I could not solve for the angle... I didn't even checked if the bullet could reach 100 m high at 90 degrees...