Projectile Motion of a Helicopter: Solving for Speed with a Falling Wheel

AI Thread Summary
A helicopter traveling at 40 m/s at a height of 100 m releases a wheel, and the goal is to determine the speed at which it hits the ground, neglecting air resistance. The initial calculations incorrectly assumed an initial vertical velocity, leading to an erroneous result. The correct approach involves using the equation for vertical motion, yielding a final vertical speed of 44 m/s just before impact. Combining this with the horizontal speed results in a total impact speed of 59 m/s. An alternative method using conservation of energy also confirms this result, highlighting the flexibility in solving the problem.
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Homework Statement



A helicopter is traveling at 40 m/s at a constant altitude of 100 m over a level field. If a wheel falls off the helicopter, with what speed will it hit the ground? Neglect air resistance.

Homework Equations



yo = 0
y = -100 m
g = -9.8 m/s^2
Vo = 0
Vxo = 40 m/s

The Attempt at a Solution



y = yo + Vot - 1/2gt^2
-100 = 0 + 0 - 1/2(9.8 m/s^2)t^2
t^2 = 100 m/(4.9 m/s^2)
=20.4 s^2
t = sqrt(20.4 s^2)
=4.5 s

y = (Vyo)t - 1/2gt^2
100 m = Vyo(4.52 s) - 1/2(9.8 m/s^2)(4.52 s)^2
(4.52 s)Vyo = 100 m + 22m
Vyo = 122 m/4.52 s
= 27 m/s

V^2 = Vx^2 + Vy^2
V = sqrt(Vx^2 + Vy^2)
= sqrt[(40 m/s)^2 + (27 m/s)^2]
= 48 m/s

Can anybody please tell me if this is correct?
 
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You solved for the time correctly although you didn't need to in order to find the final velocity in the y direction.

The second equation is wrong because the initial y velocity is 0 ; you can't solve for it. You need to use a different equation to find the velocity of the object right before it hits the ground.

Your third step would be correct if the second step was done correctly.
 
Vy^2 = Vyo^2 - 2g(y - yo)
= 0 - 2(9.8 m/s^2)(-100 m - 0)
=sqrt(1962 m^2 s^2)
= 44 m/s

therefore,

V^2 = Vx^2 + Vy^2
V = sqrt[(40 m/s)^2 + (44 m/s)^2]
= 59 m/s

Does this look better?
 
Yes, much better. That looks correct to me.
 
You could even have used ideas of work/energy to solve it, which probably would have been easier.

At the point of release, the object of mass 'm' has a known speed and height so you can work out the sum of the gravitational potential and kinetic energies. Just as it hits the ground, we know that it has zero potential energy and some unknown speed. The law of conservation of energy allows us to write the equation:

\frac{1}{2}mv_1^2 + mgh = \frac{1}{2}mv_2^2

Where we know h and v1, and the mass cancels off, allowing us to solve for v2, which gives the same answer :smile: Its a matter of preference which method you use, but i just thought id mention it as an alternative.
 
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