Question

An object is thrown vertically upward and has a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Short answer

Answer: The maximum height reached by the object is approximately 40.61 m.

Step-by-step solution

Identify the known variables

We know the following: - Speed at two thirds of maximum height, \(v = 20.0\mathrm{~m/s}\) - Fraction of the maximum height, \(h' = \dfrac{2}{3} H\) - Acceleration due to gravity, \(a = -9.81\mathrm{~m/s^2}\) (downward, hence negative) - Final velocity at maximum height, \(v_f = 0\mathrm{~m/s^2}\) Our goal is to find the maximum height, \(H\).

Use the equation of motion to find the initial velocity at two thirds height

The equation of motion that relates the velocity, the initial velocity, the height, and the acceleration is: \(v^2 = v_i^2 + 2a(h - h_i)\). At two thirds of the maximum height, the above equation becomes: \(v^2 = v_i^2 + 2a(h' - h_i)\). Since the object is thrown from the ground, we know that \(h_i = 0\). So, we have: \(v^2 = v_i^2 + 2a h'\). Solve for \(v_i\) using the given values: \((20.0\mathrm{~m/s})^2 = v_i^2 + 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).

Determine initial velocity at the ground

Rearrange the equation above to find the initial velocity: \(v_i^2 = (20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).

Calculate the final velocity at maximum height

Since the object reaches the maximum height and momentarily comes to rest, we know its final velocity, \(v_f = 0\mathrm{~m/s}\). Now we can use another equation of motion to relate the initial and final velocities, height, and acceleration: \(v_f^2 = v_i^2 + 2aH\).

Plug in the initial velocity equation to find the maximum height

Substitute the equation for \(v_i^2\) found in step 3 into the equation of motion in step 4: \(0\mathrm{~m/s^2} = ((20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H))+ 2(-9.81\mathrm{~m/s^2})H\).

Solve for the maximum height H

Now we can solve for \(H\). First, simplify the equation: \(0 = (20.0\mathrm{~m/s})^2 - (\dfrac{4}{3})(-9.81\mathrm{~m/s^2})H + 2(-9.81\mathrm{~m/s^2})H\). Next, rearrange and solve for \(H\): \(H = \dfrac{(20.0\mathrm{~m/s})^2}{(\dfrac{-2}{3})(-9.81\mathrm{~m/s^2})}\). \(H \approx 40.61\mathrm{~m}\). The maximum height reached by the object is approximately \(40.61\mathrm{~m}\).

Key concepts

Kinematics

Understanding kinematics is key when analyzing the motion of objects. Kinematics deals with the geometry of motion without reference to the forces that cause the motion. It involves several key concepts and components that define how objects move.

• Displacement: This is the change in position of an object. However, in projectile motion, we often focus on the vertical and horizontal components separately.
• Velocity: Velocity is the speed of an object in a given direction. This can change during the motion, especially under the influence of external forces such as gravity.
• Acceleration: The rate of change of velocity. In cases of objects moving within Earth's gravitational field, this is often a constant downward force denoted by the gravitational constant, approximately \(9.81 \mathrm{~m/s^2}\).

Projectile motion, like the one in our exercise, is a fascinating application of kinematic principles. Objects in projectile motion follow a curved path under the influence of gravity, where the horizontal and vertical motions are independent of each other. Grasping these essentials of kinematics is essential to understanding the complexities of projectile motion.

Equations of Motion

Equations of motion are the mathematical expressions that describe the relationship between displacement, velocity, acceleration, and time. They are essential tools in solving many problems related to kinematics.

• The first equation of motion relates velocity, initial velocity, acceleration, and time: \(v = v_i + at\). In our exercise, understanding when the final velocity becomes zero at maximum height is crucial for calculations.
• The second equation involves displacement: \(s = v_i t + \frac{1}{2} at^2\)
• The third equation is useful when time isn't known: \(v^2 = v_i^2 + 2as\). This one helps relate the velocity at different points throughout the motion, like when examining two-thirds of maximum height versus full maximum height.

Applying these equations correctly allows us to solve for one unknown if others are known. For instance, in the given exercise, we used these equations to solve for maximum height by understanding how velocity and height change throughout the object’s journey.

Maximum Height Calculation

Calculating the maximum height in projectile motion is an excellent practical application of kinematic equations. The maximum height is when the object's vertical velocity momentarily reaches zero.
In this exercise, we've already identified significant points:

• The object moves to two-thirds of its maximum height with a given speed.
• At its maximum height, vertical velocity is zero.

By using the relevant equation of motion, \(v_f^2 = v_i^2 + 2aH\), where \(v_f = 0 \, \textrm{and} \, a = -9.81 \mathrm{~m/s^2}\)\, we plug the values to determine how high the object can reach. Understanding the math behind these calculations allows one to visualize the path of the motion and the changes involved between these significant points. Listening to the gravity effects playing a dual role: it decelerates the object upwards and accelerates it downward once the peak is surpassed, this complete cycle of motion is captivating in context and concept.