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
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
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).