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A child throws a ball up with an initial velocity of 4
m/s. If the starting height of the ball was 2.5 meters,
what will be the height of the ball after 0.5 seconds?
Use a = -9.8 m/s/s.


Sagot :

The height of the ball after the given time of motion is 3.275 m.

The given parameters;

  • initial velocity of the ball, u = 4 m/s
  • height of fall of the ball, h = 2.5 m
  • time of motion, t = 0.5 s

The distance traveled by the ball is calculated as follows;

[tex]h =h_0 + ut - \frac{1}{2}gt^2 \\\\h = 2.5 + 4(0.5) - (0.5\times 9.8\times 0.5^2)\\\\h = 3.275 \ m[/tex]

Thus, the height of the ball after the given time of motion is 3.275 m.

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