V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Answer:
Moment of inertia = 0.3862kg-m²
Explanation:
2.00x10³
2.80cm
145 rad
r = r⊥ x F
F is an applied force
r⊥ is the distance between the applied force and axis
Force exerted = 2.00x10³
r⊥ = 2.8cm = 0.028m
Alpha = 145rad/s²
r = 0.028m x 2.00x10³
r = 56.0N-m
To get the moment of inertia
56.0N-m² = (145rad/s²) x I
The I would be:
I = (56.0N-m²)/(145rad/s²)
I = 56/145
= 0.3862Kg-m²
This is the moment of inertia.
Thank you!
As an object falls from rest, its gravitational energy is converted to kinetic energy
G.P.E = K.E = mgh
K.E = (80 Kg)(9.8 m/s²)(30 m)
K.E. = 23,520 J
