Question

An arrow is shot at 30 degrees above the horizontal. Its velocity is 49 m/s, and it hits the target.
a. What is the maximum height the arrow will attain?
b. The target is at the height from which the arrow was shot. how far away is it?

Answer 1

It might be helpful to use trigonometry to break up the velocity into it's x-and-y-components since there is no change in velocity in the x-direction, but gravity is changing the velocity in the y-direction.

In the case of this question, it might be easier to find out the answer to b before the answer to a, since the arrow will reach its maximum height right at the middle of its flight. However, if you've had calculus you should get an intuitive feel that the derivative of your y-position function should be equal to 0 when the arrow is at it's maximum height since that's the exact point where it's stopped going up and it's about to start going down.

I'd be happy to help more if you need to ask any questions! =D