The height need to change by 4 to double the final velocity.
<h3>Final velocity of the object</h3>
The final velocity of an object during a free fall is related to maximum height of fall as given the equation below.
v = √2gh
v² = 2gh
v²/h = 2g
v₁²/h₁ = v₂²/h₂
when v₂ = 2v₁, change in height is calculated as;
h₂ = h₁v₂²/v₁²
h₂ = (h₁ (2v₁)²) / (v₁)²
h₂ = 4h₁v₁² / v₁²
h₂ = 4h₁
Thus, the height need to change by 4 to double the final velocity.
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It is because change in the Mass of the object balances the change in the Volume of the object equally...........
Answer:

Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )


Answer:
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The answer is number (3) if am wrong am sry