You want to throw a ball so as to reach a friend 12 m away. If the desired target (your friendís 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?



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



Draw a diagram, choose a coordinate system and write all the given information in terms of the coordinates. Donít forget to assign a symbol to the unknown quantity you want to find.

I will put the origin at the throwerís 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

vyi = v0sinq , vxi = v0cosq , where v0 = 13 m/s. The coordinates are:

xi = 0, yi = 0, xf = d = 12 m, yf = 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 .



yf = 0= yi+ +vyit+1 ayt2=0+ v0sinq t-1 g t2



xf = d= xi+ +vxit=0+ v0cosq t



SOLVE Solve the problem.


t=d/ v0cosq ; v0sinq -1 gd /v0cosq =0;

2cosq sinq =sin2q =gd/ v02=(10 m/s2)(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, thatís a better choice for us because neither of us catch well, and we will have more time to judge how to catch the ball.