Gravitational potential energy, relative to some level =
(mass of the object)
times
(height above the reference level)
times
(acceleration due to gravity) .
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
Lolilolololilolollololililili
Answer:
It is neither false nor true. When they collide some of one of the objects goes to the other object.
Explanation:
<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here
then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))
The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>