Question

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?

Answer 1

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.