The velocity of the target and arrow after collision is 6.67m/s
<u>Explanation:</u>
Given:
Mass of arrow, mₐ = 415g
Speed of arrow, vₐ = 68.5m/s
Mass of the target, mₓ = 3.3kg = 3300g
speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction
Velocity of the target and arrow after collision, vₙ = ?
Applying the conservation of momentum,
mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ
415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ
28427.5 - 3630 = 3715 X vₙ
24797.5 = 3715 X vₙ
vₙ = 6.67m/s
Therefore, the velocity of the target and arrow after collision is 6.67m/s
Answer:

Explanation:
When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components


now we know that


also we know that


now plug in all data in above equations





if currents go in opposite directions, wires repel
When Object is at zero height, and there is no potential energy possess by the object then it exerts Greatest Kinetic energy in it's whole Journey
Hope this helps!
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