Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one 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 26.0 m/s. The first one is thrown at an angle of 67.5° with respect to the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first?

Answer 1

Answer:

  22.5°

Explanation:

Short answer: The trajectories will hit the same target when the projectile is launched at complementary angles. The second angle is 90° -67.5° = 22.5°.

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Longer answer: The horizontal speed of the snowball launched at angle α with speed s is ...

  vh = s·cos(α)

Then the horizontal distance at time t is ...

  x = vh·t

and the time taken to get to some distance x is ...

  t = x/vh = x/(s·cos(α))

The equation for the vertical motion of the projectile is ...

  y = -4.9t² +s·sin(α)·t

Substituting the above expression for t, we have y in terms of x:

  y = -4.9x²/(s·cos(α))² +(s·sin(α)·x)/(s·cos(α))

Factoring gives ...

  y = (x/(cos(α))(-4.9x/(s²·cos(α)) +sin(α))

The height y will be zero at x=0 and at ...

  0 = -4.9x/(s²·cos(α)) +sin(α)

  x = s²·sin(α)·cos(α)/4.9 = (s²/9.8)sin(2α)

So, for some alternate angle β, we want ...

  sin(2α) = sin(2β)

We know this will be the case for ...

  2β = 180° -2α

  β = 90° -α

The second snowball should be thrown at an angle of 90°-67.5° = 22.5° to make it hit the same point as the first.