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A volleyball is sent over a net, having been launched at a speed 5 m/s at an angle 42.7 degrees above the horizontal, from the ground. What is the ball’s speed, in m/s, 2.97 seconds after being launched?

1 answer

vx = 5cos42.7 (it's constant)
vy = 5sin42.7 - gt
Then pythagorean for total velocity

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