The cliff is 110 meters tall and the target is 21 meters away from the base of the cliff. The string is 0.55 meters long. Jim is spinning it overhead at a constant rate, holding it 1.6 meters high.

Question

The cliff is 110 meters tall and the target is 21 meters away from the base of the cliff. The string is 0.55 meters long. Jim is spinning it overhead at a constant rate, holding it 1.6 meters high.

At what angular velocity must Jim spin the rock in order for the rock to hit the target when he lets go of the string?

Answer 1

Given

Yo = 110 + 1.6 = 111.6 m

Xo = 0m X = 21m

String Length = Radius of Curvature of Rotation = 0.55m

Find: ω = ?

Solution:

Apply kinematic equations

Vertical Motion Equation of Position with respect to time under constant gravitational acceleration

Y = -(1/2)*g*t^2 + Voy*t+Yo

0 = -(1/2)*g*t^2 + Yo

t = (2*Yo/g)^1/2 = 4.769 seconds

Horizontal Motion Equation of Position with respect to time. Please note this equation can only be use

for horizontal motion since the velocity is constant

X = Vo * t + Xo

21 = Vo * 4.769

Vo = 4.40 m/s

Now to find the angular velocity of spin

Vo = ωR

Answer

ω = Vo/R = 4.4/0.55 = 8 rad/s

Solution Picture:

https://i.imgur.com/zfsAvZn.jpg