A8.1 kg object initially at rest is pushed down a 15.0 m tall hill. what is the speed of the object at the bottom of the hill?

The velocity of the object at the bottom is, v = 17.15 m/s

Explanation:

Given data,

The initial velocity of the object, u = 0

The height of the hill, h = 15 m

Let 'S' be the distance of the slope of the hill and 'Ф' be the slope of the hill formed with the ground.

The acceleration due to gravity component along the slope is given by,

                                      a = g Sin Ф

The distance of the slope since height 'h' of the hill is given,

                                       s = h / Sin Ф

Using the III equation of motion,

                                      v² = 2 as                    (∵ u = 0)

                                      v² = 2 x g Sin Ф x h / Sin Ф

                                           = 2 gh

Therefore,

                                      v = √(2gh)

Substituting the given values,

                                       v = √(2x9.8x15)

                                          = 17.15 m/s

Hence, the velocity of the object at the bottom is, v = 17.15 m/s

this means that the temperature of both substances will be equal