A juggler throws a ball from height of 0.950 m with a vertical velocity of +4.25 m/s and misses it on the way down. What is its velocity when it hits the ground?
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Check Dec,21,1:26pm Posting.
To find the velocity of the ball when it hits the ground, we can use the laws of physics, specifically the equations of motion.
We need to consider the initial height of the ball, the initial vertical velocity, and the acceleration due to gravity. The acceleration due to gravity is always directed downwards and has a value of -9.8 m/s² (negative because it acts in the opposite direction to the ball's initial motion).
The equation we can use to solve for the final velocity is:
vf² = vi² + 2ad
Where:
vf is the final velocity,
vi is the initial velocity,
a is the acceleration, and
d is the distance or displacement.
In this case, the initial velocity (vi) is +4.25 m/s (upwards), the acceleration (a) is -9.8 m/s², and the distance or displacement (d) is the initial height of the ball, which is 0.950 m.
Plugging these values into the equation, we get:
vf² = (4.25 m/s)² + 2(-9.8 m/s²)(0.950 m)
Simplifying:
vf² = 18.0625 m²/s² - 18.328 m²/s²
vf² = -0.2655 m²/s²
Since velocity cannot be negative, we discard the negative value. Thus, the square of the final velocity is 0.2655 m²/s².
Taking the square root of both sides of the equation, we find:
vf = √0.2655 m²/s²
vf ≈ 0.515 m/s (rounded to three decimal places)
Therefore, the velocity of the ball when it hits the ground is approximately 0.515 m/s.