Explanation:
Given
initial velocity(v_0)=1.72 m/s
using 
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration 
s=distance traveled
s=0.214 m
(b)time taken to travel 0.214 m
v=u+at
t=0.249 s
(c)Speed of the block at bottom
Here u=0 as it started coming downward
v=1.72 m/s
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,
Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.
v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s
Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2
Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds
Then,
x = 8*0.553 = 4.424 m
The IMA of the pulley shown is 2.
<span>The intensity of an earthquake is dependent on one's proximity to the focus of the quake, also called the "epicenter" and is based on observations of the shaking of the ground on humans, structures, and the natural landscape.</span>
