a 25 kg ball is trown into that air.when thrown it is going 10 m/s .calculate how high it travel

To calculate the height the ball will travel, we can use the equations of motion. The most appropriate equation to use in this case is the equation for vertical displacement:

Δy = (v₀ᵢ² - v₂²) / (2 * a),

where:
Δy = vertical displacement,
v₀ᵢ = initial velocity,
v₂ = final velocity,
a = acceleration.

In this case, the ball is thrown vertically upwards, so the acceleration (a) is equal to the acceleration due to gravity (-9.8 m/s²). The final velocity (v₂) is 0 m/s when the ball reaches its maximum height.

Given that the initial velocity (v₀ᵢ) is 10 m/s, we can now calculate the height. Plugging these values into the equation, we have:

Δy = (10² - 0²) / (2 * -9.8).

Δy = (100 - 0) / -19.6.

Δy = 100 / -19.6.

Δy = -5.102 m.

The negative sign indicates that the ball traveled upwards, opposite to the direction of the gravitational force. Therefore, the ball traveled approximately 5.102 meters high.

To calculate the maximum height the ball reaches, we can use the kinematic equation for vertical motion:

h = (v^2) / (2 * g)

Where:
h = height or maximum height
v = initial vertical velocity (10 m/s in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Calculating the height:
h = (10^2) / (2 * 9.8)
h = 100 / 19.6
h ≈ 5.1 meters

Therefore, the ball reaches a maximum height of approximately 5.1 meters.

V^2 = Vo^2 + 2g*h

V = 0
Vo = 10 m/s.
g = -9.8 m/s^2
Solve for h.