You need to say what the statements are if you want this answered.
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2
Find:
Horizontal displacement
Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s
Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
Hope this helps a little
initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)
<span>1.7 rad/s
The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>
