A ball is thrown vertically upward with an initial velocity of 29.4 meters per second. What is the maximum height reached by the ball?

The maximum height reached by the ball is 43.2 meters.

Using the formula:

v² = u² + 2gH

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and H is the height.

At maximum height, v = 0. Given that u = 29.4 m/s, g = -10 m/s (moving upward). Hence:

0² = 29.4² + 2(-10)H

20h = 29.4²

h = 43.2 meters

The maximum height reached by the ball is 43.2 meters.

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The maximum height reached is approximately 44.1 meters

Explanation:

The upward (vertical) velocity of the ball = 29.4 meters per second

The equation of the vertical motion of the ball under gravity, is presented as follows;

v² = u² - 2 × g × s

Where;

v = The final velocity of the ball = 0 at maximum height

u = The initial vertical velocity with which the ball was thrown= 29.4 m/s

g = The acceleration due to gravity ≈ 9.8 m/s²

s = The height reached by the ball = at maximum height

∴ 0 = 29.4² - 2 × 9.8 ×

29.4² = 2 × 9.8 ×  = 19.6·

 = 29.4³/19.6 = 864.36/19.6 = 44.1 meters

 = The maximum height reached = 44.1 meters.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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