# One strategy in a snowball fight

Here's a common college physics problem:

One strategy in a snowball fight is to throw a first snowball at a high angle over level ground. While your opponent is watching the first one, you throw a second one at a low angle and timed to arrive at your opponent before or at the same time as the first one. Assume both snowballs are thrown with a speed of 25.0 m/s. The first one is thrown at an angle of 70.0° with respect to the horizontal. (a) At what angle should the second (low-angle) snowball be thrown if it is to land at the same point as the first? (b) How many seconds later should the second snowball be thrown if it is to land at the same time as the first?

Note, this is not a homework problem for me. Solutions for this are all over the web and they can be found by searching for one strategy in a snowball fight.

Let's say point A is the initial position and point B is the final position. The final position is the same for both throws. I.e. there are two angles which result in the snowball landing in the same location. What I'm wondering is, what formula expresses this double solution?

First I'll find an expression involving th and t from one of the standard formulas for the x component of the position, velocity, and acceleration. (The notation /. (3) means "replace using equation (3)"):

            xB = xA + vxA t + 1/2 ax t^2            (1)

vxA = vA cos(th)                        (2)

xA = 0                                  (3)

ax = 0                                  (4)

(1):        xB = xA + vxA t + 1/2 ax t^2

/. (2)      xB = xA + vA cos(th) t + 1/2 ax t^2

/. (3)      xB = vA cos(th) t + 1/2 ax t^2

/. (4)      xB = vA cos(th) t                       (1.1)


Now for the y component:

            yB = yA + vyA t + 1/2 ay t^2            (5)

yA = 0                                  (6)

yB = 0                                  (7)

vyA = vA sin(th)                        (8)

(5):        yB = yA + vyA t + 1/2 ay t^2

/. (6)      yB = 0 + vyA t + 1/2 ay t^2

/. (7)      0 = 0 + vyA t + 1/2 ay t^2

/. (8)      0 = 0 + vA sin(th) t + 1/2 ay t^2

- vA sin(th) t = 1/2 ay t^2     (5.1)


So we end up with two equations and two unknowns; equations (1.1) and (5.1) with the two unknowns th and t.

I can solve those for 'th' and 't', but shouldn't those equations yield two solutions for 'th' and 't'? What am I missing?

## Update in response to answer by zhermes

Here I'll solve equation (1.1) for t and substitute that into (5.1):

(1.1):      xB = vA cos(th) t

t           xB / vA / cos(th) = t               (1.2)

(5.1):      - vA sin(th) t = 1/2 ay t^2

- vA sin(th) = 1/2 ay t

/. (1.2)    - vA sin(th) = 1/2 ay xB / vA / cos(th)

- vA^2 2 sin(th) cos(th) = ay xB

double angle formula:

- vA^2 sin(2 th) = ay xB

sin(2 th) = - ay xB / vA^2          (5.2)

th = arcsin(- ay xB / vA^2) / 2     (5.3)


Looking at equation (5.2), yes, it's clear that since sin(2 th) is symmetrical about 45 degrees, there will be two answers if th is in (0, 45) or (45, 90).

• Your second equation has two solutions for $t$, one of which is $t=0$, because the sonwball is at ground level when you throw it, and again when it reaches $x_B$. So you only need to consider the non-zero solution. Commented Dec 7, 2012 at 1:01

If you solve for $t$ in Eq. (5.1), and plug that into equation (1.1), you'll see that the solution looks like $x_B \propto v_A^2 sin(\theta) cos(\theta)$. The function on the right is symmetric about $\pi/4$, thus, as long as $\theta$ doesn't equal $\pi/4$, there will be two solutions (symmetrically about $\pi/4$). Of course, in general, there could be one, two, or no solutions (see figure below).