A 180 pound body is dropped from a helicopter 5 miles (26400 ft.) up -
- what maximum speed would the body attain, in miles per hour, before hitting the ground?
Disregard gravity, air density, etc. factors, a very rough approximation is all that is needed. Thanks.
The question is meaningless ignoring gravity.
In fact it is hard to say ignoring air drag as well.
Because the body would presumably reach constant terminal velocity where air drag up exactly balances gravitational force down.
Damon,
Where do you suggest I look for an approximate mph figure.
I realize we are ignoring key factors, but surely someone can give me an educated guess - -
e.g. the body would hit the ground at 100 mph plus or minus 10 mph.
Thanks
To calculate the maximum speed a body would attain when dropped from a height, we can make use of the principle of conservation of energy.
The potential energy (PE) of the body when it is initially dropped is given by the product of its weight (W) and the height (h) from which it is dropped. In this case, the weight is 180 pounds and the height is 5 miles (or 26400 ft).
PE = W * h
PE = 180 lbs * 26400 ft
Next, we equate the potential energy to the kinetic energy (KE) of the body just before it hits the ground. The kinetic energy is given by the formula:
KE = (1/2) * m * v^2
In this case, the mass (m) of the body can be obtained from its weight using the conversion factor that 1 pound is equal to 32.174 pounds (this is approximately equal to the acceleration due to gravity).
m = W / g
where g is the acceleration due to gravity (32.174 ft/s^2).
Substituting the value of m in the kinetic energy equation, we have:
KE = (1/2) * (W/g) * v^2
Now, equating the potential energy to the kinetic energy, we can solve for the velocity (v):
PE = KE
W * h = (1/2) * (W/g) * v^2
Simplifying the equation, we find:
h = (1/2) * (v^2/g)
Rearranging the equation to solve for v, we get:
v = sqrt(2 * g * h)
Substituting the values of g (32.174 ft/s^2) and h (26400 ft) into the formula, we can calculate the maximum speed attained:
v = sqrt(2 * 32.174 ft/s^2 * 26400 ft)
v ≈ 1982 ft/s
Now, to convert this value to miles per hour, we multiply it by the conversion factor of 3600 seconds per hour and 1 mile per 5280 feet:
v_mph = (1982 ft/s * 3600 s/h) / 5280 ft/mi
v_mph ≈ 1353 mph
Therefore, in this rough approximation, the maximum speed attained by the body before hitting the ground is approximately 1353 mph.