The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time
The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.
Distance = 10 * 1.06
Distance = 10.6 meters
The boat should be 10.6 meters away horizontally from the point of release.
<span>(v1)^2=(v0)^2-2ad; v1= final velocity; v0= initial velocity; a=gravitational constant;d=distance
Since the stone was dropped we know v0=0.
(v1)^2=2ad
(136)^2=2(32)d
d=(136)^2/(2*32)
d=289ft</span>
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For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied, the speed of the 10-kg block is mathematically given as
V=4.33m/s
<h3>What is the speed of the 10-kg block?</h3>
Generally, the equation for the workdone is mathematically given as
W=T tehta<dtheta
W=100(90*\pi/180)
W=1.5707*100
W=1.57Nm
The change in potential energy across the pulley
dP=mgh
dp=10*9.81*111.8
dp=10.97J
For the thrid position, potential energy is
dP=mg(0.3)
dP=17.658J
dP'=17.658J-13.125
dP'=-4.532J
For 2nd position dP=0
The change in Kinectic energy across the pulley\
dK.E=0.5mv^2
For 1st
dk.E=0.5m(10)^2
2nd
dK.E=0.5Iw^2
dK.E=0.5(7.5*10^-3)(v^2/0.05)^2
3rd
2nd=3rd
In conclusion,
157.07=dKE+dP.E
157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53
V=4.33m/s
Read more about Kinetic energy
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