Question

A 17614kg helicopter moves upward with a constant acceleration from rest to a height of 407m. The spinning blades create an upward thrust of 316447N that lifts the helicopter into the air. How fast is the helicopter moving as it reaches a height of 407m?

a. v=77.574 m/s
b. v=80.524 m/s
c. v=85.784 m/s
d. v=71.874 m/s
e. None of these is correct.

Answer 1

The speed of the helicopter as it reaches a height of 407m is approximately 85.784 m/s. Option C is correct.

Step-by-step explanation:

The speed of the helicopter can be calculated using the kinematic equation:

v^2 = u^2 + 2as

Where,

v = final velocity (unknown)

u = initial velocity (0 m/s since the helicopter starts from rest)

a = acceleration (unknown)

s = distance (407 m)

Plugging in the values, we get:

v^2 = 0^2 + 2 * a * 407

Since the helicopter starts from rest, the initial velocity (u) is 0, and the equation simplifies to:

v^2 = 814a

Next, we can use Newton's second law to find the acceleration:

F = ma

The upward thrust of the helicopter is equal to the weight of the helicopter:

316447N = 17614kg * g

(Where g = acceleration due to gravity = 9.8 m/s^2)

Solving for the mass (m), we get:

m = 316447N / 9.8 m/s^2

Next, we can substitute the value of m into the equation for acceleration:

a = 316447N / (9.8 m/s^2 * 17614kg)

Finally, we can substitute the value of a back into the equation for velocity to find the speed at a height of 407 m:

v^2 = 814 * (316447N / (9.8 m/s^2 * 17614kg))

Taking the square root of both sides, we get:

v = sqrt(814 * (316447N / (9.8 m/s^2 * 17614kg)))

Calculating this expression gives us a result of approximately 85.784 m/s.

Therefore, the correct option is (c) v = 85.784 m/s.