The problem states that a ball is dropped from rest and falls under the influence of gravity (neglecting air resistance). At the same time, another ball is thrown with an initial velocity of v0 from a point that is 10 meters horizontally away and 5 meters below the first ball (see figure in the problem statement). The question is asking us to determine the angle θ at which the second ball should be thrown so that it hits the first ball.


A. theta = 45°

B. theta = 60°

C. theta = 90°

D. Depends on v0.

I have been thinking about this problem and I have still not grasped how it is 60°.

This is the figure that was asked for, no floor.

enter image description here

So someone in a thread said that we can assume the initial velocity is 0.1×c, meaning that the initial velocity is not important but the angle theta is. So how do we come to the conclusion that theta = 60° from tan (theta) = 10/2, exactly? More importantly, where does the 2 in "10/2" come from?

t any rate, tan(theta) = 5 gives us that tan⁻¹ (5) = 1.37π which is 246°, which is just one period + 66°, not sure how that can be rounded down to 60° anyway.

This was all supposed to be solved without a calculator by the way. This question is of a multiple choice character, so we can reason that C, 90°, must be false since we would only land in the same place if we assume no air resistance. So we're left with A, B. D was false.

  • $\begingroup$ What is the source of the problem? Is this the figure provided with the original text of the problem? $\endgroup$
    – nasu
    Apr 28, 2023 at 3:40

3 Answers 3


You have to aim at the ball before it starts moving, than both balls have the same acceleration downwards so they must meet, if v0 is large enough so they meet before the falling ball hits the ground .

  • $\begingroup$ Okay but how do we get the angle $$\theta = 60°$$, more importantly, how do we do that without having access to a calculator during the exam? $\endgroup$
    – cricket900
    Apr 7, 2023 at 18:17
  • $\begingroup$ Since you aim at the ball the angle is 60° or 30° to horizontal an usually you know sin, cos tan for the angles 0,30,45,60 or you can calulate it $\endgroup$
    – trula
    Apr 8, 2023 at 16:17
  • $\begingroup$ But the angle is not excatly 60° only about $\endgroup$
    – trula
    Apr 8, 2023 at 17:19

This is the classic monkey and hunter problem in disguise. The answer is that the "thrown" ball must be aimed directly at the initial position of the "target" ball. As the balls fall, gravity will pull them both the same distance downward relative to the positions they would have had if there was no gravity. And so no matter how strong gravity is, the ball will hit the target.

Once you realize this, you should also realize that the angle of 60° is actually incorrect, and that in fact none of the given options are correct.

Also, for fun: Here's the demonstration done on a large scale by the good folks at the University of Minnesota.


If the balls must collide, they must encounter themselves at the same position and, in this case at same study time. So,

y-coord.(ball 1) = y-coord.(ball 2)

$5\;{\rm{m}}- gt^2/2 = v_0 t \cos{\theta}- gt^2/2$

$5\;{\rm{m}} = v_0 t\cos{\theta} $ (1)


x-coord.(ball 1) = x-coord.(ball 2)

$10\;{\rm{m}}= v_0 t \sin{\theta}$ (2)

Taking "$v_0 t$" term from (2)and putting it into (1), or dividing both equations member by member, it is obtained:

$1/2 = \cot{\theta}$


$\tan{\theta} = 2$

$\theta \simeq 63.4^{\circ}$

  • $\begingroup$ If you look this result $\tan{\theta}=2$, automatically, items A, C and D are neglected without calculator. Firstly, because $\tan(45)=1$, second, $\tan(90)=\infty$ and third, the term $v_0$ doesn't appear in the result. $\endgroup$
    – LuisDFQ
    Apr 28, 2023 at 1:59
  • $\begingroup$ But, you take in account that all the options are incorrect, 63.4 degrees doesn't round to 60. $\endgroup$
    – LuisDFQ
    Apr 28, 2023 at 2:05

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