Answer to Question #96107 in Classical Mechanics for Layla

Question #96107
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 ◦
1
Expert's answer
2019-10-08T10:04:45-0400

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.


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