Question

Students at an engineering contest use a compressed air cannon to shoot a soft ball being hoisted straight up at 10m/s by a crane.The cannon tilted upward at an angle of 30 degrees is 100m from the box and fires by remote control the instant the box leaves the ground.Students can control the launch speed of the soft ball by setting the air pressure.What launch speed should the students use to hit the box?

Answer 1

you still there?

Answer 2

So this is a simple problem involving equations of motion. You can treat the horizontal and vertical components independently.
You can use the following: \[s = ut + \frac{at^2}{2} \] where \(s\) is the displacement (not distance - important difference!), \(u\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time elapsed.

In this case, I would solve for \(t\) using the horizontal information (Initial velocity is 0m/s, distance is 100m). Once you've done that you can solve for the initial vertical speed \(u\) - note that the total displacement is 0 as you return to the same height that you left.

You can then combine the two speeds using pythagoras. Note that there are two contributions to vertical velocity that you will need to consider!

Good luck!