If a rock climber accidentally drops a 56.0 g piton from a height of 295 m, what would its speed be just before striking the ground? ignore the effects of air resistance.

The speed of the piton before hitting the ground = 76.04 m/s

Explanation:

From the law of conservation of energy,

The kinetic energy of the piton  = potential energy of the piton

Note:Before striking the ground, the potential energy of the piton is converted to kinetic energy, with no energy lost to air resistance.

1/2mv² = mgh Equation 1

Where m = mass of the piton, v = speed of the piton, h = height, g = acceleration due to gravity.

making v the subject of the equation above,

v =√(2gh) Equation 2

Given: h = 295 m, g = 9.8 m/s²

Substituting into equation 2

v = √(2×9.8×295)

v = √5782

v = 76.04 m/s

Thus the speed of the piton before hitting the ground = 76.04 m/s

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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