Answer: 
Explanation:
If we make an analysis of the net force 
of the rock that was thrown upwards, we will have the following:
(1)
Where:
is the force with which the rock was thrown
is the weight of the rock
Being the weight the relation between the mass 
of the rock and the acceleration due gravity 
:
(2)
(3)
Substituting (3) in (1):
(4)
(5) This is the net Force on the rock
On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:
(6)
Finding the acceleration 
:
(7)
(8)
Finally:
Answer:
547 m
Explanation:
From law of motion
s = ut + ½at²
Where "t" is Time taken to reach Earth
s= distance= 182 m
a= vertical acceleration = 5.82 m / s 2
U= initial velocity in vertical position = 0
182= ½ × 5.82t²
t²=( 2× 182)/ 5.82
= 364/5.82
= 62.54
t= √62.54
t= 7.908s
horizontal distance travelled = speed x time
Horizontal speed= 72.6 m / s
horizontal distance travelled =72.6× 7.908
= 547 m
Hence, the survivor will it hit the waves at 547 m away
Answer:
there yah go that's the answer
Answer:
a) 
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂)
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂) 2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ -
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘= - u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system
