Answer to Question #96107 in Classical Mechanics for Layla

Question

One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
While your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
Assume both snowballs are thrown with
the same initial speed 34.7 m/s. The first
snowball is thrown at an angle of 66◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of ◦

Answer 1

Let us write the equations of motion of a snowball, thrown at angle $\theta$, with initial speed $v_0$:

$x = v_0 cos \theta \cdot t$ , $y = y_0 + v_0 \sin \theta \cdot t - \frac{g t^2}{2}$ .

When the snowball reaches the target, the height is the same as initial, hence $y_0 = y_0 + v_0 sin \theta \cdot t - \frac{g t^2}{2}$ , from where $v_0 \sin \theta = \frac{g t}{2}$, and $t = \frac{2 v_0 sin \theta}{g}$. Plugging in the last expression into equation for $x$, obtain $x = \frac{v_0^2}{g} 2\sin \theta \cos \theta = \frac{v_0^2}{g} \sin 2 \theta$ (using trigonometric identity $\sin 2 x = 2 \sin x \cos x$).

If we have two snowballs, thrown at different angles $\theta_1, \theta_2$ with the same initial speed $v_0$, their horizontal positions must be equal , hence $\frac{v_0^2}{g} \sin 2 \theta_1 = \frac{v_0^2}{g}\sin 2 \theta_2$, from where $\sin 2 \theta_1 = \sin 2 \theta_2$. This condition must hold for two angles, to implement out snowball fight strategy. The condition holds when $\theta_2 = \frac{\pi}{2} - \theta_1$, so we need to throw the second snowball at an angle $90^{\circ} - 66^{\circ} =24^{\circ}$ above the horizontal.