Question

A boy is whirling around a stone tied at the end of a 0.71 m long string at constant angular speed of 4 revolutions per second in a plane perpendicular to the ground. The string breaks when it is making an angle of 54 degrees above the horizontal and the stone is on the way up. The stone is 0.93 m above the ground when this happens. How long does the stone take to hit the ground after the string breaks?

Answer 1

"a stone tied at the end of a 0.71 m long string"
r=0.71 m

"4 revolutions per second"
ω=4*2π radians/s

Tangential speed, v0
=rω

"The stone is 0.93 m above the ground"
Distance above ground, h
= 0.93

"an angle of 54 degrees above the horizontal"
θ=54°

Vertical component of initial velocity, vv
=v0sinθ

Let t=time it takes to hit ground,
S=vv*t+(1/2)gt²
-0.93 = vv*t+(1/2)(-9.8)t²

Solve for t.
I get -0.051 sec. or 4 sec. approx.
We reject -0.051 because t>0.

Answer 2

I tried doing the problem your way, but I got confused at the end.
For the equation S = w*t + (1/2)(-9.8)t**2, what was the value of w?
When I tried using w = v0 (17.844m/s), I got the same answers as yours
But when I tried using w = v0sin54 (14.4363m/s), I got different answers. I'm not sure which way is right. I'm leaning toward the latter one.

Answer 3

You're quite right.
I have calculated for 5 revolutions per second. That is why they don't match your answers, which are on the right track.
I get now -0.06 and 3 seconds, and -0.06 is to be rejected.
What do you get for your answers?
Sorry for the inconvenience.

Answer 4

Yes, I also got -0.06 and 3.00 seconds, so obviously the answer is 3.00 seconds.
Thanks for your help :)

Answer 5

You're welcome!