A player serves a volleyball striking the volleyball at a 60 degree angle with a velocity of 15 m/s.What is the horizontal velocity of the volleyball 0.5 seconds before it achieves max height?
Question
A player serves a volleyball striking the volleyball at a 60 degree angle with a velocity of 15 m/s.What is the horizontal velocity of the volleyball 0.5 seconds before it achieves max height?
Solution
To solve this problem, we need to understand that the horizontal velocity of a projectile (in this case, the volleyball) remains constant throughout its flight, assuming no air resistance. This is due to the fact that gravity only acts vertically, not horizontally.
The horizontal velocity (Vx) can be calculated using the initial velocity (V) and the angle (θ) at which the ball was hit. The formula is:
Vx = V * cos(θ)
Given that the initial velocity (V) is 15 m/s and the angle (θ) is 60 degrees, we can substitute these values into the formula:
Vx = 15 m/s * cos(60)
The cosine of 60 degrees is 0.5, so:
Vx = 15 m/s * 0.5 = 7.5 m/s
Therefore, the horizontal velocity of the volleyball 0.5 seconds before it achieves max height is 7.5 m/s.
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