Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
We have that the instantaneous velocity of the shuttlecock when it hits the ground is
From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as
Where
acceleration is still -9.81 m/s2,
Hence,
Therefore
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