While sitting in a math class, you become curious and decide to put the math and your physics to work in a practical way. If you can shoot a spit-wad from a straw with a total speed of 6 m/s, at what angle would you have to shoot it in order for it to hit a blackboard that is 1 m from you, at a height that is even with your mouth? Keep in mind a trig identity: 2sin theta cos theta = sin2theta.

Question

While sitting in a math class, you become curious and decide to put the math and your physics to work in a practical way.  If you can shoot a spit-wad from a straw with a total speed of 6 m/s, at what angle would you have to shoot it in order for it to hit a blackboard that is 1 m from you, at a height that is even with your mouth?  Keep in mind a trig identity:  2sin theta cos theta  =  sin2theta.

First off, why does it even have to be pointed at an angle (as opposed to straight ahead), and what is the approach to finding the angle once I've drawn it out, considering the velocity?

bobpursley · Answer

find time in air:
in the vertical direction:
hf=hi+6sinTheta*t-4.9t^2
but hf=hi=0, so t= 6sinTheta/4.9
in the horizontal direction:
distance=velocity*time
1m=6cosTheta*6sinTheta/4.9
Now you can use the angle formula you were given.

Steve · Answer

the article in wikipedia on Trajectory explains this well. And of course there has to be an angle. It takes some time for the spit wad to get to the target, and it drops during that interval ...