Two dimensional motion problem (kicking a soccer ball into the goal)

AI Thread Summary
The discussion focuses on solving a two-dimensional motion problem involving kicking a soccer ball into a goal. The user initially applies the formulas for the x and y components of motion but realizes discrepancies in their calculations, particularly regarding the angle of projection. A correction is suggested for the x component equation, emphasizing the need to consider the total flight time and ascent time. Additionally, the importance of using the correct gravitational constant is highlighted, as it affects the calculations. The conversation concludes with a reminder to utilize the formulas for range and total time of flight to solve the problem accurately.
Ineedhelpwithphysics
Messages
43
Reaction score
7
Homework Statement
Carli Lloyd scored half-field goal (53 m) vs. Japan with a beautiful projectile motion. It looks from the video that the ball was inthe air for 2.0 seconds. What was the angle of the ball when it left her foot?
Relevant Equations
VoX = vcos(theta)
VoY = vsin(theta)
Vf(y) = Vo(y) + a(y)t
Delta y = VoY + 1/2a(y)t^2
I did this too fast idk if I'm wrong

So for the x component
use the formula d = rt / delta x = v*t
26 = vcos(theta)

y component
use the displacement formula
19.6 = vsin(theta)

tan^-1(19.6/26) = 36.5 degrees
the answer key says 20.2 degrees idk whats wrong
 
Physics news on Phys.org
I did wrong calculations please ignore this
 
I get the angle 20.67 with ##g=10m/s^2##.
Your first equation is almost correct, it should be $$\frac{53}{2}=v_0\cos\theta$$
But no clue how did you infer the 2nd equation.
Hint: if the total flight time is 2.0 sec what is the total ascent time, ##t_A## ,that is the time the ball rises from the initial point to its highest point, and how ##v_{0y}=v_0\sin\theta## relates to ##t_A##?

Now I see that your second equation is also almost..... correct up to a factor of 2 e hehe and with ##g=9.8m/s^2##
 
What is the formula the for range ##R## of a projectile and its total time of flight ##T##? Those two are the ones you need to answer the question @Ineedhelpwithphysics
 
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