I have to do a physics project where we pick a scene from a movie and decide whether the physics used were good or bad. I chose the scene in Get Him to the Greek where a character falls into a pool but hits his arm on the concrete side and breaks it. I need to show equations that prove that the fall was high enough to break his arm.
The fall lasted exactly 3 seconds. The force needed to break an arm is 73 newtons. gravity = 9.8 m/s^2. The actor weighs 161 lbs or 73 kg
so far I have: acceleration = ((9.8 x 3.0s) - 0) / 2 = 14.7 m/s^2
And Force = mass x acceleration: F = 73 kgs x 14.7 m/s^2 = 1073.1 N
Conclusion: Because the force needed to break an arm is 73 newtons, and the force with which he hit the concrete is 1073 newtons, the fall was more than enough to break his arm.
Is there anything wrong with my math?
The fall lasted exactly 3 seconds. The force needed to break an arm is 73 newtons. gravity = 9.8 m/s^2. The actor weighs 161 lbs or 73 kg
so far I have: acceleration = ((9.8 x 3.0s) - 0) / 2 = 14.7 m/s^2
And Force = mass x acceleration: F = 73 kgs x 14.7 m/s^2 = 1073.1 N
Conclusion: Because the force needed to break an arm is 73 newtons, and the force with which he hit the concrete is 1073 newtons, the fall was more than enough to break his arm.
Is there anything wrong with my math?
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Your acceleration in the fall is way too high. At most it will be about g = 9.8 m/s^2, which is the acceleration due to gravity near Earth's surface. So right away your numbers are suspect.
The next issue is M = 73 kg. That's the mass of the person, not of the arm. If we assume an arm mass m = M/10 = 7.3 kg that's the mass striking the poolside and receiving the force F. It would be M if the arm were an integral part of the total mass, but it is hinged and will give on impact, so not all the mass of the person is behind that impact on the arm.
To find the average impact force F = m V/dT = m sqrt(2gH)/dT = 7.3*sqrt(2*9.8*(1/2)*9.8*3^2)/1 = 214.6 Newtons or higher, assuming an impact interval dT = 1 sec. H = 1/2 gt^2 where t = 3 sec, is the height from which the person fell. V is the impact speed assuming the person started with U = 0 initial speed at H. [See source.]
The 215 N > 73 N should break the hapless victim's arm.
NOTE: The impact interval is highly uncertain. For example, had the person fallen into a pile of leaves along side the pool, they would have cushioned the fall and dT might be greater than 1 second. If the impact interval were extended to dT = 3 sec instead of 1 sec, that would lower the force to just less than 73 N and the arm might not break. A pile of leaves would not do the job, but a big air bag might (which is what stunt men use). But if he fell on cement and on top of his own arm, dT could very well have been 1/2 a second or less. The point is this; what he fell onto and the attitude he was in when impacting the side of the pool will have a bearing on dT and consequently F.
The next issue is M = 73 kg. That's the mass of the person, not of the arm. If we assume an arm mass m = M/10 = 7.3 kg that's the mass striking the poolside and receiving the force F. It would be M if the arm were an integral part of the total mass, but it is hinged and will give on impact, so not all the mass of the person is behind that impact on the arm.
To find the average impact force F = m V/dT = m sqrt(2gH)/dT = 7.3*sqrt(2*9.8*(1/2)*9.8*3^2)/1 = 214.6 Newtons or higher, assuming an impact interval dT = 1 sec. H = 1/2 gt^2 where t = 3 sec, is the height from which the person fell. V is the impact speed assuming the person started with U = 0 initial speed at H. [See source.]
The 215 N > 73 N should break the hapless victim's arm.
NOTE: The impact interval is highly uncertain. For example, had the person fallen into a pile of leaves along side the pool, they would have cushioned the fall and dT might be greater than 1 second. If the impact interval were extended to dT = 3 sec instead of 1 sec, that would lower the force to just less than 73 N and the arm might not break. A pile of leaves would not do the job, but a big air bag might (which is what stunt men use). But if he fell on cement and on top of his own arm, dT could very well have been 1/2 a second or less. The point is this; what he fell onto and the attitude he was in when impacting the side of the pool will have a bearing on dT and consequently F.