One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released from the circle at the angle θ so that it will hit the target. The angle in the drawing is 33.4°. The distance to the target from the center of the circle is d. (See the drawing below, which is not to scale.) The circular path is parallel to the ground, and the target lies in the plane of the circle. The distance d is two times the radius r. Ignore the effect of gravity in pulling the stone downward after it is released and find the angle θ.

Question

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Answer 1

I did it all out for you in my reply below.

Answer 2

So what is the solution?

Answer 3

Once again - You did not scroll down to see my full reply
Here it comes again

Answer 4

The only physics in this problem is that the pebble continues at its velocity straight as it was tangent to the circle after release since there is no longer any force accelerating it toward the center of the circle. The rest is geometry.

Pebble at A, theta degrees from x axis

Target at B, (180 - P) degrees from x axis. P = 33.4

center at C as drawn

Draw tangent to circle from B to A

Then CA is perpendicular to BA

so
ABC is a right triangle with right angle at A

CA = R
CB = 2R

so call angle BCA Z
cos Z = R/2R = 1/2 so Z = 60 degrees
so
theta = 180 - 33.4 - 60
= 86.6

Answer 5

Thanks!!!

Answer 6

Thank you! It's very helpful.

Answer 7

Thanks! This really helps.

Answer 8

no

Answer 9

This is not the answer in my book.
The book answer is 61 degrees