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A man throws a ball up with a velocity of 30 m/s. How high does it get?

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vf =
vi =
t =
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User Sebastiaan
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1 Answer

7 votes

Final answer:

To calculate the height the ball reaches when thrown up with a velocity of 30 m/s, we can use the kinematic equation vf^2 = vi^2 + 2aΔy. Plugging in the values, we find that the ball reaches a height of approximately 45.92 meters.

Step-by-step explanation:

In order to calculate the height the ball reaches, we need to use the kinematic equation:

vf^2 = vi^2 + 2aΔy

Where:

  • vf is the final velocity, which is 0 m/s because the ball is at rest at the highest point.
  • vi is the initial velocity, which is 30 m/s because the ball is thrown up with that velocity.
  • a is the acceleration due to gravity, which is approximately 9.8 m/s^2.
  • Δy is the change in height, which is what we want to find.

Plugging in the values, we get:

0 = (30)^2 + 2(-9.8)Δy

Simplifying the equation, we have:

900 = -19.6Δy

Dividing both sides by -19.6, we find:

Δy ≈ -45.92 m

Therefore, the ball reaches a height of approximately 45.92 meters.

User FireFoxII
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