Question

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). What maximum height above the projection point does it reach?

Short answer

Answer: The maximum height reached by the object is approximately 39.98 meters.

Step-by-step solution

List the known variables

We are given the initial velocity (\(v_0\)) of the object which is \(28.0 \mathrm{~m} / \mathrm{s}\). We also know that the acceleration due to gravity is \(-9.81 \mathrm{~m} / \mathrm{s}^{2}\), and we can set the initial and maximum heights to be \(0\). Lastly, the final velocity (\(v\)) will be \(0 \mathrm{~m} / \mathrm{s}\) when the object reaches its maximum height.

Choose the appropriate kinematic equation

We will use the equation \(v^2 = v_0^2 + 2 * a * (h - h_0)\), where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(h - h_0\) represents the change in height. In our case, \(v = 0 \mathrm{~m} / \mathrm{s}\), \(v_0 = 28.0 \mathrm{~m} / \mathrm{s}\), \(a = -9.81 \mathrm{~m} / \mathrm{s}^{2}\), and \(h_0 = 0\).

Solve for the maximum height (h)

Using the equation from step 2, we can solve for \(h\): \(0^2 = (28.0 \mathrm{~m} / \mathrm{s})^2 + 2 * (-9.81 \mathrm{~m} / \mathrm{s}^2) * (h - 0)\) Now, solve for \(h\): \((0 - (28.0 \mathrm{~m} / \mathrm{s})^2) / (2 * (-9.81 \mathrm{~m} / \mathrm{s}^2)) = h\) \((-28^2 \mathrm{~m^2} / \mathrm{s^2}) / (2(-9.81 \mathrm{~m} / \mathrm{s}^2)) = h\) \(h \approx 39.98~m\) So, the maximum height above the projection point that the object reaches is approximately \(39.98\) meters.

Key concepts

Kinematic Equations

To understand the behavior of objects in motion, physicists rely on kinematic equations. These equations describe the relationships between displacement, time, velocity, and acceleration in motion with constant acceleration. One of the foundational kinematic equations, and particularly crucial for projectile motion, is given by \(v^2 = v_0^2 + 2a(h - h_0)\).

In this context, \(v\) represents the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(h - h_0\) is the change in height. When calculating the maximum height of a projectile, we set the final velocity to zero because at its peak the projectile’s vertical velocity momentarily stops before coming down due to gravity. By rearranging the kinematic equation to solve for height \(h\), we are able to determine the peak of the object's trajectory.

Initial Velocity

The initial velocity \(v_0\) of an object is the speed at which it starts its movement. In the context of projectile motion, initial velocity can have both horizontal and vertical components. Since we are focused on the maximum height, only the vertical component of the initial velocity is relevant for our calculations.

In the given exercise, the initial velocity is a known value of \(28.0 \text{ m/s}\) directed upward. This value is crucial to determining the maximum height that the projectile will reach. The larger the initial velocity, the higher the object will go before the force of gravity brings it back down.

Acceleration Due to Gravity

All objects near the earth's surface experience acceleration due to gravity, which is approximately \(9.81 \text{ m/s}^2\) directed downwards. This acceleration is denoted as \(-g\) in equations when considering the upward motion as positive.

Gravity is a constant force that influences how quickly an object will return to the ground after being projected upwards. In our equation, it appears as a negative value because it opposes the initial upward velocity. Understanding this acceleration is essential for predicting the maximum height and overall trajectory of projectiles.

Projectile Motion Physics

Projectile motion physics involves studying the motion of objects that are launched into the air and are subject to gravity. It is important to consider that in projectile motion, the horizontal and vertical components of the motion are independent of each other.

In our problem, we focus on the vertical component because we’re interested in how high the object goes. Assuming no air resistance, the only force acting on the projectile is gravity, acting downward. The motion can be described by kinematic equations. At the maximum height, the vertical velocity is zero, and from that point, the projectile starts to descend back to its original height or another level. Maximum height is a key aspect in analyzing projectile motion, and understanding it is fundamental for many applications in physics and engineering.