Question

A ball with mass 0.15 kg is thrown upward with initial velocity 10 m/s from the roof of a building 100 m high. Assume there is a force due to air resistance of magnitude |v|/30 directed opposite to the velocity, where the velocity v is measured in m/s.

Find the maximum height above the ground that the ball reaches.

Answer 1

The maximum height above the ground that the ball reaches is approximately 104.61 meters.

Step-by-step explanation:

To find the maximum height, we can use the kinematic equations of motion. Considering the forces acting on the ball, including air resistance, we use the following formula:

`\[ h_{\text{max}} = h_0 + (v_0^2)/(2g) - (1)/(2)\left((v_0)/(30)\right)^2 \]`

where:

-
`\( h_{\text{max}} \)`is the maximum height,

-
`\( h_0 \)` is the initial height (100 meters in this case),

-
`\( v_0 \)` is the initial velocity (10 m/s upward),

- g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the formula, we get:

`\[ h_{\text{max}} = 100 + (10^2)/(2 * 9.8) - (1)/(2)\left((10)/(30)\right)^2 \approx 104.61 \]`

Therefore, the maximum height above the ground that the ball reaches is approximately 104.61 meters. The calculation takes into account both the initial conditions and the force of air resistance.