A baseball pitcher makes a big mistake and throws the ball straight up. If it reaches a maximum height of 15 m find the initial velocity he threw it up with

17.15 m/s.

Explanation:

The following data were obtained from the question:

Maximum height (h) = 15 m

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going again gravity)

Initial velocity (u) =.?

The initial velocity with which the baseball pitcher threw the ball can be obtained as follow:

v² = u² + 2gh

0² = u² + (2 × –9.8 × 15)

0 = u² + (–294)

0 = u² – 294

Collect like terms

0 + 294 = u²

294 = u²

Take the square root of both side

u = √294

u = 17.15 m/s

Therefore, the baseball pitcher threw the ball with an initial velocity of 17.15 m/s

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