Gravitational potential energy of baseball

AI Thread Summary
The discussion revolves around calculating the speed of a baseball thrown from a 22.0-meter building using energy conservation principles. For part (a), the initial potential energy converts to kinetic energy just before impact, allowing the calculation of speed without needing the mass of the baseball. In part (b), the speed remains the same regardless of the throw angle due to energy conservation. Participants clarify that mass can be omitted in the calculations since it cancels out, simplifying the problem. The key takeaway is that both scenarios yield the same final speed when air resistance is ignored.
aftermatth
Messages
2
Reaction score
0

Homework Statement


A baseball is thrown from the roof of a 22.0-m-tall building with an inital velocty of magnitude 12.0 m/s and directed at an angle of 53.1 degrees above the horizontal.
a) What is the speed o fthe ball just before it strikes the ground? Use energy conservation and ignore air resistance.
b) What is the answer for part (a) if the inital velocity is at an angle of 53.1 below the horizontal?
c) If the effects of air resistance are included, will part (a) or part (b) give the higher speed?

Vo = 12 m/s
y1 = 22m
y2 = 0m
a = -9.8 m/s²
Voy=(12m/s)sin53.1 = 9.6 m/s
don't know why they don't give any variables for mass of the baseball.. ;\
other problems use .145kg for baseball so i decided just to use that

Homework Equations


E= K + Ugrav = constant
K= 1/2mv1²
1/2mv1² + mgy1 = 1/2mv2² + mgy2

The Attempt at a Solution


Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.

K1 + Ugrav1 = K2 + Ugrav2
0 + Ugrav1 = K2 + 0
Ugrav1 = K2 (initial gravitational potential engergy = final kinetic energy)
K2 = 1/2mv2² = (1/2)(.145kg)(9.6m/s)² = 6.7J
Ugrav1 = mgy1 = (.145kg)(9.8m/s²)(22m) = 31.2 J
and I'm stuck.

how do i find the speed?

for part (b) i know that the velocity will be the same as the first part, because when it returns back from throw point, it will end up having the same velocity as if it were thrown 53.1 degrees above.
 
Last edited:
Physics news on Phys.org
aftermatth said:
Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.
You are given the initial speed of the ball, so it has kinetic energy.
 
Incidentally, the mass of the ball is NOT needed to solve this problem.

Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy.

Does the ball only have potential energy to start with? Isn't it launched at a certain speed?
 
Oops, sorry for repeating what Doc Al said
 
Doc Al said:
You are given the initial speed of the ball, so it has kinetic energy.

oh youre right it does have kinetic energy..
K1 + Ugrav1 = K2

mass isn't required? do i just omit it from all parts?
i still don't know how to get to the velocity.
 
Last edited:
You will find if you divide both sides of:
K1 + Ugrav1 = K2
by the mass, the problem does not contain mass.
 
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