How Do You Calculate the Average Force of Air Resistance on a Projectile?

AI Thread Summary
To calculate the average force of air resistance on a projectile, one must first determine the difference in potential energy (PE) between the maximum height without air resistance and the actual height reached. The projectile, with a mass of 0.656 kg and an initial speed of 25.4 m/s, theoretically reaches 32.9 m without air resistance, while it only ascends to 7.64 m in reality. The work done by air resistance can be calculated using the difference in PE, which is approximately 49.116032 J. This work is then divided by the distance traveled (7.64 m) to find the average force of air resistance, resulting in a value of about 21.3 N. Understanding these calculations is crucial for accurately determining the effects of air resistance on projectile motion.
neoncrazy101
Messages
22
Reaction score
0

Homework Statement


A projectile of mass 0.656 kg is shot straight up with an initial speed of 25.4 m/s. (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only 7.64 m, determine the magnitude of the average force due to air resistance.

Homework Equations


1/2mvf^2 + mghf = 1/2mvo^2 + mgho

Work = KE+ PE
W = Fcos(90)s

The Attempt at a Solution



Ok, I am able to get the answer for A: (32.9m) but with B, I cannot get it. What should I do to get the answer for B? Whenever I put the equation into W = Fcos(90)s I come up with zero which is not the answer. So I don't know what I'm doing wrong.
 
Physics news on Phys.org
Well, cos(90) = 0 (some sort of trig identity I suppose), so your equation for the work done by the air resistance is wrong. Looking at the problem, the projectile is shot straight up, and the air resistance acts to retard this motion. The reduction in trajectory height would be similar to using the same effort to launch a heavier projectile upward.
 
neoncrazy101 said:
Ok, I am able to get the answer for A: (32.9m) but with B, I cannot get it. What should I do to get the answer for B? Whenever I put the equation into W = Fcos(90)s I come up with zero which is not the answer. So I don't know what I'm doing wrong.
What is the work done by air resistance? (hint: what is the difference in maximum potential energy between A and B?).

Now if that work is done by a constant force over the distance that the projectile moves, what is that force?

AM
 
The difference in PE between A and B is 49.116032.

I'm probably very very confused (most likely, brain is fried plus headache) but if I take that 49.116032 and multiply it by the 7.64m then I come with up 375.2 which is really large.

EDIT: I figured it out! W=162.50248 (roughly) and then F = W/D thus 162.50248/7.64 = 21.3ish.
 
Last edited:
neoncrazy101 said:
The difference in PE between A and B is 49.116032.

I'm probably very very confused (most likely, brain is fried plus headache) but if I take that 49.116032 and multiply it by the 7.64m then I come with up 375.2 which is really large.

EDIT: I figured it out! W=162.50248 (roughly) and then F = W/D thus 162.50248/7.64 = 21.3ish.
How did you get this value for W? (you should show your work). What are the units of force?

AM
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top