# Projectile Motion helicopter

1. Sep 28, 2008

### Mitchtwitchita

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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 any body please tell me if this is correct?

2. Sep 28, 2008

### Mattowander

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.

3. Sep 28, 2008

### Mitchtwitchita

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?

4. Sep 28, 2008

### Mattowander

Yes, much better. That looks correct to me.

5. Sep 28, 2008

### danago

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 Its a matter of preference which method you use, but i just thought id mention it as an alternative.