You throw a baseball straight up in the air with an initial velocity of 41 m/s What is the maximum height the baseball reaches above your hand?

85.77 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 41 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

The maximum height reached by the ball can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 41² – (2 × 9.8 × h)

0 = 1681 – 19.6h

Collect like terms

0 – 1681 = –19.6h

–1681 = –19.6h

Divide both side by –19.6

h = –1681 / –19.6

h = 85.77 m

Thus, the maximum height reached by the ball is 85.77 m

answer; equivalent to a one(1) horsepower motor;