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.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The value is 
Explanation:
From the question we are told that
The length of the stretcher is 
The weight of the stretcher is 
The weight for Wayne is 
The distance of center of gravity for Wayne from Chris is 
Generally taking moment about the first end where Chris is
=> upward moment
Here
is the force applied by Jamie
Generally taking moment about the second end where Jamie is
=> downward moment
Generally at equilibrium , the upward moment is equal to the downward moment

=> 
=> 
Newtons second law.. <span>The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.</span>
Answer:
F = 0.535 N
Explanation:
Let's use the concepts of energy, at the highest and lowest point of the trajectory
Higher
Em₀ = U = mg y
Lower
= K = ½ m v²
Emo =
mg y = ½ m v2
v = √ 2gy
y = L - L cos θ
v = √ (2g L (1-cos θ))
Now let's use Newton's second law n at the lowest point where the acceleration is centripetal
F = ma
a = v² / r
In turning radius is the cable length r = L
F = m 2g (1-cos θ)
Let's calculate
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N