What Angle and Timing Adjustments Allow Two Thrown Snowballs to Collide?

  • Thread starter Thread starter svtec
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
Two snowballs are thrown in the same direction, one at a 70-degree angle and the other at a lower angle, both with a speed of 25.0 m/s. The discussion centers on determining the angle for the lower snowball to ensure both collide at the same point and the timing for its throw. There is confusion regarding the final velocity of the snowballs upon impact, with clarification needed on using uniform acceleration equations to calculate the time and horizontal distance traveled. The importance of distinguishing between the moment before collision and the final stationary velocity is emphasized. The conversation highlights the challenges faced by beginners in physics and the need for clear understanding of projectile motion concepts.
svtec
Messages
22
Reaction score
0
In a snowball fight 2 snowballs are thrown in the same direction, but one is thrown 70.0 degrees with respect to the horizontal and the other is thrown some arbitrary distance lower. they both are thrown with a speed of 25.0 m/s.

At what angle should the one that is thrown lower be to arrive at the same point as the one that was thrown at 70.0 degrees?

How many seconds later should the lower thrown snowball be thrown to arrive at the same time as the higher one.

resolving into x and y components for the first one i get.

x initial = 0
x final =?
v initial = 25.0cos(70.0)
v final = 0 m/s
a = 0 m/s^2

y initial = 0
y final = ?
v initial = 25.0sin(70.0)
v final = 0m/s
a = -9.80m/s^2

then i get lost...


tia...


-andrew
 
Physics news on Phys.org
What makes you think "v final = 0 m/s"?
 
i figured that when the snowball made contact with the target it's velocity would be 0.
 
As someone who has been hit often by snowballs before, I can reassure you that as the ball hits you, it is definitely not stationary. (You are confused with collision modelling here, I think. The velocity after it hit is presumeably zero. But not as it hits/just before it hits.)

What you want to do is to use the uniform acceleration equations in each case to work out an expression for time taken in each case for the ball to fall to the ground, and then insert this to find out the horizontal distance traveled for each, and equate the two distances to find the angle.
 
i am kinda confused. this is my first quarter in physics and i seem to confuse the moment before collision and the stationary component of the final velocity. i guess that is why i was enlisting the help of people that are far more superior in their understanding of physics then myself...

i'll give what you said a shot.

thanks...

-andrew
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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