Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s
Acceleration = ▵v/▵t
Time = d/v
Fisrt calculate time : ( 118/29 ) = 4 seconds
Then calculate acceleration
A = 29/4 = 7.25 m/s²
Now the force.
Force = mass * acceleration.
F= 1,019 * 7.25
F= 7,387 N
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:

where F is the force applied, and

is the stretch of the spring with respect to its equilibrium position. Using the data, we find

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
Answer and Explanation:
distance will be 2×3.14 (pie)×r
displacement will be 2r (diameter)
the motion is uniform circular motion as the object is moving in a circular path with uniform motion