Is this even solvable? Kinematics.

  • Thread starter Thread starter tophat
  • Start date Start date
  • Tags Tags
    even Kinematics
AI Thread Summary
The discussion centers on solving a kinematics problem involving a ball thrown vertically upward, requiring the determination of its maximum height, h, and initial velocity, V0, given that the total motion lasts 10 seconds. Participants emphasize the importance of using the correct final velocity just before the ball lands on the building, rather than at the point of stopping. They suggest employing kinematic equations that relate distance, time, and acceleration due to gravity. A numerical solution for both V0 and h is confirmed to exist, necessitating algebraic manipulation and logical reasoning. The problem is deemed solvable with the right approach to the equations.
tophat
Messages
1
Reaction score
0

Homework Statement



A ball is thrown vertically upward from ground level with an initial velocity V0. The ball rises to a height, h, then lands on the roof of a building of height 1/2h. The entire motion requires 10s. Find the height, h, and the initial velocity, V0.

Homework Equations



Kinematics.

The Attempt at a Solution



I can't come up with anything, I'm using a final velocity of 0 m/s when running everything through and none of the values are making sense. I just need to know if it's actually solvable for real values. I'm in a hurry for work right now, but I will check back later.

I'm using: t2: 10s
t0= 0s
v1= 0 m/s (turning point)
a = -g
y1= h
y2= 1/2h
v2= 0 m/s (stop)
y0 = 0 m
 
Physics news on Phys.org
I think using that as your final velocity is your main problem so far. Yes, the final velocity will be zero once the ball hits the building and stops, but you can't use that value because the process of the object stopping isn't free-fall motion. You have to use the value of the velocity just before it lands as the final velocity.
 
You've just jotted down some trivial equations so far; you haven't actually written the proper kinematic equations involving distance as a function of time.

Yes, there is a numerical solution for V0 and h, and there is a fair amount of algebra and logical deduction involved in obtaining it.
 
s(t)=v_0t-1/2gt^2, s(t)_{\text{max}}=h=f(v_0,g), solve s(t)=\frac{h}{2} for t, and then plugin t=10s, get the larger answer
 
Last edited:
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