While walking past the on going construction site at the location of Zachry, you

Question

While walking past the on going construction site at the location of Zachry, you stop to observe a 1000 tonne crane lowering a 80 kg construction worker at a constant downward velocity of 16.7 m/s . The worker throws a 1.0-kg stone with an initial velocity of 15.9 m/sperpendicular to the path of the descending crane (i.e. as per the worker's view point, he has thrown the stone horizontally). The worker sees the stone hit the ground 13.1 s after it was thrown. Assume that over the entire motion the crane continues to descend downward at the same constant speed of 16.7 m/s . Assume that the ground is flat, i.e. the base/foot of the crane is at the same vertical level as the vertical level where the stone lands.

a) How high was the crane, as seen by you, when the stone was thrown? Express your answer in base SI units.

b) How high is the crane (or the worker in the crane) above ground level when the stone hits the ground? Express your answer in base SI units.

c) How far is the stone from the worker when the stone hits the ground? Express your answer in base SI units.

d) Just before the stone hits the ground, what are its horizontal and vertical velocity components as measured by the worker in the crane?

e) Just before the stone hits the ground, what are its horizontal and vertical velocity components as measured by you who is at rest on the ground?

Explanation / Answer

given,

downward velocity = 16.7 m/s

mass of worker= 80 kg

mass of stone = 1 kg

velocity of the stone = 15.9 m/s

time = 13.1 sec

vertical distance covered in 13.1 sec = 16.7 * 13.1 + 0.5 * 9.8 * 13.1^2

hight of the crane, when the stone was thrown = 1059.659 m

distance covered by the crane in 13.1 sec = 16.7 * 13.1

distance covered by the crane in 13.1 sec = 218.77 m

hight of the crane (or the worker in the crane) above ground level when the stone hits the ground = 1059.659 - 218.77

hight of the crane (or the worker in the crane) above ground level when the stone hits the ground = 840.889 m

horizontal distance covered in 13.1 sec = 15.9 * 13.1

distance to the stone from the worker = 208.29 m

horizontal component of the velocity will remain same as no force is acting horizontally so,

horizontal component of the velocity = 15.9 m/s

by first equation of motion

v = u + at

v = 16.7 + 9.8 * 13.1

v = 145.08 m/s

vertical component of velocity = 145.08 m/s

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