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2 Dec 2021
Problem 66
Page 73
Section: General Problems
Chapter 3: Kinematics In Two Dimensions; Vectors
Textbook ExpertVerified Tutor
2 Dec 2021
Given information
Net height is .
Net is from the server.
Ball is launched from a height of .
Step-by-step explanation
Step 1.
From the second equation of motion
Here is height, is the vertical component of velocity, is acceleration due to gravity and is time
As the ball is launched from a height of and the ball has to clear the net of height . So the height we take in the formula is the difference of these two.
So now the height is
The vertical component of velocity will be zero , as the ball goes horizontally.
So here
Thus substituting the values we get
(equation )
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