Question

ball with mass 0.15 kg is thrown upward with an inital velocity 20 m/s from the roof of a building 30 m high. there is a force due to air resistance of |v|/30, 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 20.36 meters.

To find the maximum height above the ground that the ball reaches, you can use the following steps:

Step 1: Calculate the acceleration due to gravity (g).

The acceleration due to gravity near the surface of the Earth is approximately 9.81 m/s² (assuming standard conditions).

Step 2: Calculate the net force acting on the ball.

The net force on the ball is the difference between the force due to gravity (weight) and the force due to air resistance. Since the ball is thrown upward, the weight of the ball acts downward, and the air resistance opposes the upward motion.

Weight (W) = mass (m) * gravity (g)

W = 0.15 kg * 9.81 m/s² ≈ 1.47 N

The force due to air resistance (F_air) is given as |v|/30, where v is the velocity in m/s.

Step 3: Calculate the velocity at the maximum height.

At the maximum height, the velocity of the ball momentarily becomes zero. So, set |v|/30 equal to zero and solve for v:

|v|/30 = 0

|v| = 0

This means that at the maximum height, the velocity is zero.

Step 4: Calculate the initial velocity of the ball when it reaches the maximum height.

The ball was thrown upward with an initial velocity of 20 m/s.

Step 5: Use the kinematic equation to find the maximum height (h) above the ground.

The kinematic equation for finding the maximum height reached by an object in free fall is:

`\[v_f^2 = v_i^2 + 2as\]`

Where:

vf is the final velocity (0 m/s at the maximum height).

vi is the initial velocity (20 m/s).

a is the acceleration (negative due to gravity, -9.81 m/s²).

s is the displacement (height above the ground).

Rearrange the equation to solve for s:

`\[0 = (20 m/s)^2 + 2*(-9.81 m/s²)*s\]`

`\[0 = 400 - 19.62s\]`

Now, solve for(s:

`\[19.62s = 400\]`

`\[s = 400 / 19.62 \approx 20.36 \, \text{m}\]`