0
$\begingroup$

need help with a problem I created.

enter image description here

Someone is running at 18mph west, toward B from C, the distance between A and C is 0.2 miles Someone shoots a paintball that travels at 190 mph, from A toward B, predicting where C would be. At what angle would A have to shoot? From A to C is 90 degrees. Assume time starts right after the gun is shot

How do I get the angle (degrees) that the gun should be shot to hit the target? (no sway, air resistance, etc)

Would this be enough information to get a angle? Any help would be appreciated I can change the way I setup the problem

$\endgroup$
1
$\begingroup$

In order for the paintball and the runner to be in the same place at the same time, you require the component of velocity of the paintball perpendicular to the initial displacement to be the same as that of the runner. Therefore you require $$190\sin \theta=18$$ Note that the actual initial distance $0.2$ is not relevant. All this assumes velocities are constant, and they set off simultaneously.

$\endgroup$
  • $\begingroup$ need help understanding something, If both the target and the ball travel at 190 mph, then arcsin(190/190) = 90 (degrees) the ball is launched, but then they would never touch each other? In other words the ball will never be able to reach the target because of the distance? I think I got it. thanks. $\endgroup$ – hit Sep 13 '15 at 14:30
  • $\begingroup$ Yes that's right. The target has to be moving slower than the projectile for there to be a hit in this situation. $\endgroup$ – David Quinn Sep 13 '15 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.