physics
posted by kia .
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 10.0 m/s. The first one is thrown at an angle of 65.0° with respect to the horizontal.
(a) At what angle should the second (low angle) snowball be thrown to arrive at the same point as the first?

The vertical velocity component of the first snowball is 10sin65 = 9.063m/s.
The time to impact derives from Vf = Vo  9.8t or 0 = 9.063/9.8 = .9248 sec.
The distance to the target derives from d = V^2(sin2µ)/g where V = the initial velocity of the snowball, µ = the angle of the velocity vector to the horizontal and g = the acceleration due to gravity or d = 10^2(sin130)/9.8 = 7.816 meters.
For the second snowball to hit the same target, it must be thrown at an angle of 25º to the horizontal producing
d = 10^2(sin50)/9.8 = 7.816 meters.
Being thrown at a shallower angle, it would reach the target sooner as derived from 0 = Vv  9.8t = 10sin25  9.8t making t = .4312 seconds.
Therefore, to reach the target at the same time as the first snowball, it must be thrown .4935 seconds after the first.