You want to throw a ball so as to reach a friend 12 m away. If the desired target (your friend’s hand) is at the same height from which you launch the ball, and you can throw at 13 m/s, at what angle should you throw the ball?

MODEL

I will model the ball as a projectile moving in a plane. I will neglect air resistance.

SETUP

Draw a diagram, choose a coordinate system and write all the given information in terms of the coordinates. Don’t forget to assign a symbol to the unknown quantity you want to find.

I will put the origin at the thrower’s hand, the *x*-axis horizontal toward the friend, and the *y*-axis vertically upward. The angle I am trying to find is measured from the horizontal and is called q
. The velocity components at launch are

*v _{yi }*= v

*x _{i}* = 0,

Next decide how you will solve the problem, and write down the equations you plan to use.

I will solve for the time that the projectile is in the air. The answer will depend on q . Then I will put the time in the x-equation and solve for q .

*y _{f}* = 0=

*x _{f}* = d=

SOLVE Solve the problem.

t=d/ v* _{0}*cosq

2cosq
sinq
=sin2q
=gd/ v_{0}^{2}=(10 m/s^{2})(12 m)/(13 m/s)^{2}=0.71 (no units - they all cancel)

Thus 2q = 45° or 135°

q =22.5° or 67.5°

ANALYZE Discuss your result

There are two possible angles. The ball will take longer to reach my friend if I use the greater angle, that’s a better choice for us because neither of us catch well, and we will have more time to judge how to catch the ball.