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

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.

imageshack.us/photo/my-images/39/physic.gif/

Answer 1

So what is the solution?

Answer 2

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

Answer 3

This is not the answer in my book.

The book answer is 61 degrees

Answer 4

To find the angle θ, we can use the properties of a circle and the given information.

First, let's understand the problem. A slingshot with a pebble is whirled on a circular path with radius r. The pebble is released at an angle θ and we need to find this angle.

Since the circular path is parallel to the ground and the target lies in the plane of the circle, we can assume that the line connecting the center of the circle to the pebble is perpendicular to the ground.

From the given information, we know that the distance to the target from the center of the circle is d, which is twice the radius r.

Now, we have a right triangle formed by the line connecting the center of the circle to the pebble, the horizontal distance (r), and the vertical distance (d).

Using trigonometry, we can relate the angles and sides of this triangle. In this case, we need to find the angle θ, which is the angle between the line connecting the center to the pebble and the radius.

We have the opposite side (d), and we need to find the adjacent side (r) to use the tangent function.

Tangent of an angle is defined as the ratio of the opposite side to the adjacent side. So, we can use the tangent function to find the angle θ.

tan(θ) = opposite/adjacent
tan(θ) = d/r

Since d is twice the radius r, we can substitute d with 2r:

tan(θ) = (2r)/r
tan(θ) = 2

Now, we need to find θ. We can take the inverse tangent (also known as arctan) of both sides of the equation to find the angle:

θ = arctan(2)

Using a calculator, we find that arctan(2) is approximately 63.4°.

Therefore, the angle θ is approximately 63.4°.

Answer 5

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

Here it comes again

Answer 6

Thank you! It's very helpful.

Answer 7

Thanks! This really helps.

Answer 8

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