Question

A ball with mass \(0.15 \mathrm{kg}\) is thrown upward with initial velocity \(20 \mathrm{m} / \mathrm{sec}\) from the roof of a building \(30 \mathrm{m}\) high. Neglect air resistance. $$ \begin{array}{l}{\text { (a) Find the maximum height above the ground that the ball reaches. }} \\ {\text { (b) Assuming that the ball misses the building on the way down, find the time that it hits }} \\ {\text { the ground. }} \\\ {\text { (c) Plot the graphs of velocity and position versus time. }}\end{array} $$

Short answer

Answer: The maximum height above the ground that the ball reaches is approximately 50.2 m, and it takes approximately 4.4 seconds for the ball to hit the ground.

Step-by-step solution

Identify the knowns and unknowns

In this problem, we have: - Mass of the ball, \(m = 0.15 \, \mathrm{kg}\) - Initial velocity, \(v_0 = 20 \, \mathrm{m/s}\) - Height of the building, \(h = 30 \, \mathrm{m}\) - Acceleration due to gravity, \(g = 9.81 \, \mathrm{m/s^2}\) We are supposed to find: - Maximum height above the ground (part a) - Time it takes for the ball to hit the ground (part b) - Plot the graphs of velocity and position versus time (part c)

Calculate the maximum height above the ground (part a)

We can find the maximum height by using the following kinematic equation: \(v^2 = v_0^2 - 2g(y - y_0)\) At the maximum height, the final velocity \(v\) will be 0. Also, \(y_0 = h\), the initial height. Therefore, the equation becomes: \(0 = v_0^2 - 2g(y_{max} - h)\) Solving for the maximum height, \(y_{max}\): \(y_{max} = h + \frac{v_0^2}{2g}\) Plug in the known values and calculate \(y_{max}\): \(y_{max} = 30 + \frac{(20)^2}{2(9.81)} \approx 50.2 \, \mathrm{m}\) So, the maximum height above the ground that the ball reaches is approximately 50.2 m.

Calculate the time it takes for the ball to hit the ground (part b)

To find the time it takes for the ball to hit the ground, we'll use the following kinematic equation: \(y = y_0 + v_0t - \frac{1}{2}gt^2\) When the ball hits the ground, \(y = 0\). So, the equation becomes: \(0 = h + v_0t - \frac{1}{2}gt^2\) We can solve this quadratic equation for \(t\): \(t^2 - \frac{2v_0}{g}t + \frac{2h}{g} = 0\) Plug in the known values and solve for \(t\) using the quadratic formula: \(t^2 - \frac{2(20)}{9.81}t + \frac{2(30)}{9.81} = 0\) Solving this quadratic equation, we get two roots, but we're only interested in the positive root, which is approximately 4.4 seconds. So, the time it takes for the ball to hit the ground is approximately 4.4 seconds.

Plot the graphs of velocity and position versus time (part c)

To plot the graphs of velocity and position as functions of time, we'll use the following kinematic equations: \(y(t) = h + v_0t - \frac{1}{2}gt^2\) and \(v(t) = v_0 - gt\) For the position-time graph, use the equation \(y(t) = 30 + 20t - \frac{1}{2}(9.81)t^2\), and plot it for the interval \(t = 0\) to \(t = 4.4 \, \mathrm{sec}\). For the velocity-time graph, use the equation \(v(t) = 20 - 9.81t\), and plot it for the same time interval. These graphs will show how the position and velocity of the ball evolve over time during its trajectory.

Key concepts

Maximum Height in Projectile Motion

When studying the motion of projectiles, a frequently explored parameter is the maximum height that the projectile reaches. This is the peak of its trajectory, where the projectile's vertical velocity becomes zero before descending back to the ground.

In the context of our example where a ball is thrown upward with a given initial velocity from a certain height, the maximum height can be calculated using a kinematic equation which relates the final velocity (which is zero at the maximum height), the initial velocity, the acceleration (due to gravity, in this case), and the height difference.

The formula to determine the maximum height (y_{max}) above the starting position is given by the relation:
\[y_{max} = y_0 + \frac{v_0^2}{2g}\]
Here, y_0 is the initial height from which the ball is thrown (the height of the building), v_0 is the initial velocity of the ball, and g is the acceleration due to gravity. By substituting the known values into this equation, we can find the highest point that the ball reaches.

Time of Flight

The time of flight of a projectile refers to the total time the projectile spends in the air from the moment it is launched until the moment it returns to the ground. To find the time of flight for our ball being thrown from the rooftop, we use a quadratic equation derived from the kinematics of uniformly accelerated motion:

\[0 = h + v_0t - \frac{1}{2}gt^2\]
By setting the final vertical position y to zero (the ground level), and rearranging this equation, we can solve for t, which represents the time it takes for the ball to reach the ground after being thrown.

Using the quadratic formula and disregarding the negative root (since time cannot be negative), we obtain the time at which the ball makes impact. In our example problem, we determined this time to be approximately 4.4 seconds.

Position-Time Graph

A position-time graph represents how the position of an object changes over time. In projectile motion, such a graph can reveal key characteristics of the object's trajectory, such as its maximum height and the time of flight.

For the ball thrown upward from the building, the position-time graph is a parabola that opens downwards due to the negative acceleration (gravity) acting on the ball. The equation for the position as a function of time is:
\[y(t) = y_0 + v_0t - \frac{1}{2}gt^2\]
This equation allows us to plot the trajectory, showing the ascent and descent of the ball, with the peak of the parabola indicating the highest point reached by the ball. By plotting this equation from t = 0 to the time of flight, we can visually assess the entire journey of the ball.

Velocity-Time Graph

A velocity-time graph demonstrates how the velocity of an object changes as time progresses. In the case of our vertical projectile motion, the graph will be a straight line with a negative slope, reflecting constant acceleration due to gravity.

The velocity-time relationship can be expressed by:
\[v(t) = v_0 - gt\]
When plotted, the line starts from the initial velocity v_0 at t = 0 and decreases linearly due to the negative term -gt. The point where the graph crosses the time axis gives the time it took the ball to reach the maximum height, after which the velocity of the ball becomes negative, showing it is moving downwards. Plotting this graph for the same time interval as the position-time graph will also indicate when the ball hits the ground, as it will match with the time of flight obtained earlier.