Question

A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. Neglect air resistance.

a. Find the maximum height above the ground that the ball reaches.
b. Assuming that the ball misses the building on the way down, find the time that it hits the ground.
c. Plot the graphs of velocity and position versus time.

Answer 1

The maximum height above the ground that the ball reaches is 20.41 m. The ball hits the ground after approximately 5.98 seconds. Equations for velocity and position versus time are provided to plot the graphs.

Step-by-step explanation:

a. To find the maximum height the ball reaches, we can use the equation of motion for vertical motion, which is given by:

Final velocity squared = Initial velocity squared + 2 * acceleration * displacement

The acceleration is -9.8 m/s^2 because the ball is moving against the force of gravity. The initial velocity is 20 m/s, and the displacement is the maximum height, which is 30 m. Plugging these values into the equation, we can solve for the final velocity, which will be 0 m/s at the maximum height. Rearranging the equation, we get:

Displacement = (Final velocity squared - Initial velocity squared) / (2 * acceleration)

Substituting the values, we get:

Displacement = (0 - 20^2) / (2 * -9.8)

Displacement = -400 / -19.6

Displacement = 20.41 m

Therefore, the maximum height above the ground that the ball reaches is 20.41 m.

b. To find the time that the ball hits the ground, we can use the equation of motion for vertical motion again. Since the ball is thrown upward and then comes back down, the total displacement will be twice the height of the building, which is 60 m. The initial velocity is 20 m/s, and the acceleration is -9.8 m/s^2. Plugging these values into the equation, we get:

Displacement = Initial velocity * time + (1/2) * acceleration * time^2

Substituting the values, we get:

60 = 20 * time + (1/2)(-9.8) * time^2

Simplifying the equation:

4.9 * time^2 + 20 * time - 60 = 0

Using the quadratic equation to solve for time, we get:

time = (-20 ± sqrt(20^2 - 4 * 4.9 * -60)) / (2 * 4.9)

After solving the equation, we get two possible values for time: 5.98 seconds and -2.50 seconds. Since time cannot be negative in this context, the ball hits the ground after approximately 5.98 seconds.

c. To plot the graphs of velocity and position versus time, we can use the equations of motion for vertical motion. The equation for velocity is given by:

Velocity = Initial velocity + acceleration * time

Substituting the values, we get:

Velocity = 20 - 9.8 * time

The equation for position is given by:

Position = Initial position + Initial velocity * time + (1/2) * acceleration * time^2

Substituting the values, we get:

Position = 30 + 20 * time - (1/2) * 9.8 * time^2

Answer 2

(a) The maximum height reached by the ball is 20.4 m

(b) The time taken for the ball to hit the ground is 3.21 s.

(c) The graph of position and velocity versus time is shown in the image attached.

How to calculate the maximum height reached by the ball?

(a) The maximum height reached by the ball is calculated by applying the following formula.

v² = u² - 2gh

where;

• v is the final velocity
• u is the initial velocity
• g is acceleration due to gravity
• h is the height reached by the ball

at maximum height, v = 0

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (20² ) / (2 x 9.8)

h = 20.4 m

(b) The time taken for the ball to hit the ground is calculated as;

t = √ (2H/g)

where;

• H is the total height of fall of the ball
• g is acceleration due to gravity

H = 20.4 m + 30 m

H = 50.4 m

t = √ (2 x 50.4 / 9.8)

t = 3.21 s