A helicopter descends to the ground at a constant speed of 20 m/s. When it reach

Question

A helicopter descends to the ground at a constant speed of 20 m/s. When it reaches a height of 25 m, a wrench slips off the edge of the helicopter floor near an open door. At this point the wrench moves vertically under the influence of just the Earth's gravity. As observed from the ground, what is the initial acceleration of the wrench the instant just after the wrench slips out the door of the helicopter? Give the a. With what speed will the wrench hit the ground? From the moment the wrench leaves the helicopter, what is the time interval it takes for the wrench to hit the ground? b. c.

Explanation / Answer

here,

initial vertical speed , u = 20 m/s

heigth , h0 = 25 m

a)

the initial accelration of the wrench is accelration due to gravity i.e g = 9.81 m/s^2 acting downwards

b)

let the final speed be v

using conservation of energy

0.5 * m * v^2 = m * g * h

v = sqrt(2 * 9.81 * 25) m/s

v = 22.1 m/s

c)

let the time taken be t

v= u + g * t

22.1 = 20 + 9.81 * t

t = 0.21 s

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