Concept: Air Resistance in Projectile Motion

AI Thread Summary
Air resistance significantly impacts the horizontal range of a projectile, such as a tennis ball, more than its maximum height due to the differing effects of velocity on drag forces. At a 45-degree launch angle, the horizontal velocity remains constant, while the vertical velocity decreases as the ball rises and falls, leading to a greater average horizontal speed being affected by air resistance. The concept of increased drag at higher speeds explains why the horizontal distance is reduced more than the vertical height, as the force of air resistance increases with velocity. Additionally, the average vertical velocity is lower than the horizontal velocity throughout the trajectory, contributing to the disparity in the effects of air resistance. Understanding these dynamics clarifies why horizontal motion is more influenced by air resistance than vertical motion in projectile motion scenarios.
ibwm
Messages
4
Reaction score
0
Hi, I am in need for some physics help regarding the effect of air resistance on projectile motion.

My question is "For a tennis ball stuck at an angle of 45 degrees to the horizontal at typical speed without spin, why does air resistance have a greater effect on the horizontal range of the ball rather than the maximum height of the ball?"

I've tried searching Google for a solution, but have yet been able to find one. I don't just want to copy the answer off of the internet, but actually learn the concept.

Any help is much appreciated,
Sharna.

Edit:
In my notes I have:

For launch angle of 45 degrees, Vh and Vv = sin45 and cos45 = 0.707N (I have no idea why it says Newtons here if it's talking about velocity)

And then my notes say: "Theory: Vh= constant = x. Vv= x --> 0 --> x. Average Vh = x (more friction), Average Vv=x/2 (less friction)"

I am still unsure as to what all of this means. What is the reasoning behind the horizontal range being affected more by air resistance than the vertical height? Does it have something to do with friction forces being larger when velocity is larger?
 
Last edited:
Physics news on Phys.org
Here is a hint: in which case do you feel more wind? Driving at 10mph? Or driving at 60mph?
 
DaveC426913 said:
Here is a hint: in which case do you feel more wind? Driving at 10mph? Or driving at 60mph?

You'd feel more wind driving at 60mph, yeah? I know that there's more friction the faster you are travelling, but I am unsure as to how this affects horizontal distance more than vertical distance. What is the concept behind this?
 
ibwm said:
You'd feel more wind driving at 60mph, yeah? I know that there's more friction the faster you are travelling, but I am unsure as to how this affects horizontal distance more than vertical distance. What is the concept behind this?
What is the average velocity of the ball in the y direction as compared the average velocity of the ball in the x direction? They definitely be can't the same since at one point in its trajectory, y velocity is zero.
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top