The Graph above shows the speed of a car traveling in a straight line as a function of time. The car accelerates uniformly and r
1 answer:
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Answer:
vₓ = 20 m/s, v_{y} = -15 m / s
Explanation:
This is a conservation of moment problem, since it is a vector quantity we can work each axis independently
The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved
X axis
Initial moment. Before the crash
p₀ = m₁ v₀ₓ + m₂ v₀ₓ
Final moment. After the crash
p_{fx} = (m₁ + m₂) vₓ
p₀ₓ =
m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ
vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)
vₓ = v₀ₓ = 20 m/s
Y Axis
Initial
p_{oy} = m₁ v_{oy}
Final
p_{fy} = (m₁ + m₂) v_{y}
p_{oy} = p_{fy}
the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s
m₁ = (m₁ + m₂) v_{y}
v_{y} = m₁ / (m₁ + m₂) v_{oy}
v_{y} = 1/4 60
v_{y} = -15 m / s
Vertical speed is down
Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:
where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,
<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:
where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,
Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:
where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,
<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:
where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,
<u>vf = 107.91 m/s</u>