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 object reaches a maximum height of approximately 12.24 m above the launch point.

Step-by-step solution

Identify given information and required variables

Here's what we know: - The object is thrown vertically upward, so the acceleration is due to gravity: \(a = -9.8 \mathrm{~m/s^2}\) - The speed when the object reaches two thirds of its maximum height: \(v = 20.0 \mathrm{~m/s}\) The variable we need to find is the maximum height, \(H\).

Apply the kinematic equation for the two-thirds height

Let's write the equation for the two thirds of the maximum height: $$v^2 = u^2 + 2a(\frac{2}{3}H)$$ We don't know the initial velocity, \(u\), so let's get this equation in terms of \(u\). $$u^2 = v^2 - 2a(\frac{2}{3}H)$$

Apply the kinematic equation for the maximum height

Now, let's apply the kinematic equation for the maximum height. At this point, the object starts to fall, so the final velocity, \(v\), is 0: $$0^2 = u^2 + 2aH$$

Solve for H in terms of u

Now, we have two equations in terms of \(u\): 1. \(u^2 = v^2 - 2a(\frac{2}{3}H)\) 2. \(u^2 = -2aH\) By setting the expressions for \(u^2\) equal to each other, we can solve for \(H\): $$v^2 - 2a(\frac{2}{3}H) = -2aH$$

Solve for the maximum height H

Now we can solve for \(H\): $$H = \frac{v^2}{2a} - \frac{2}{3}H$$ Multiplying both sides by 3 and rearranging the terms, we get: $$5H = \frac{3v^2}{2a}$$ Finally, we can solve for the maximum height \(H\): $$H = \frac{3v^2}{10a}$$

Plug in values and find H

Now, we can plug in the known values to find the maximum height: $$H = \frac{3(20.0 \mathrm{~m/s})^2}{10(-9.8 \mathrm{~m/s^2})} = \frac{1200 \mathrm{~m^2/s^2}}{98 \mathrm{~m/s^2}} \approx 12.24 \mathrm{~m}$$ Thus, the object reaches a maximum height of approximately \(12.24 \mathrm{~m}\) above the launch point.

Key concepts

Kinematic Equations

Understanding the motion of objects requires a fundamental grasp of kinematic equations – the mathematical formulas that describe the motion of objects under the influence of forces such as gravity. In physics, these equations are crucial for analyzing how objects move in various situations, including projectile motion.

The kinematic equations relate the variables of an object's motion: its initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), time (\(t\)), and displacement (\(x\) or \(h\), depending on the context). Here's the list of these equations, assuming constant acceleration:

• \( v = u + at \) (final velocity)
• \( x = ut + \frac{1}{2}at^2 \) (displacement)
• \( v^2 = u^2 + 2ax \) (relates velocity and displacement without time)
• \( x = v t - \frac{1}{2}at^2 \) (displacement when initial velocity is zero)

These equations are applied to solve various physics problems, including finding the maximum height of projectile motion, by eliminating variables that aren't known or aren't required for the particular solution. In our original problem, we mainly used the third equation to calculate the maximum height by relating it to the velocity and acceleration at two-thirds of the maximum height. This kinematic equation, tailored to a projectile's vertical motion, allows us to summarize a projectile's journey without knowing the time involved.

Vertical Projectile Motion

In vertical projectile motion, objects move under the sole influence of gravity, assuming air resistance is negligible. This type of motion follows a symmetrical path – the time to ascend to the maximum height is equal to the time of descent from that height. Two key stages define this motion:

• The upward journey, where the object slows down until it reaches its peak.
• The downward journey, where the object speeds up as it falls back to its original launch point or another surface.

In our exercise, these principles help to deduce that at the maximum height, the final velocity (\(v\)) is 0, as gravity has decelerated the object to a momentary halt before it starts descending. Using the symmetry of projectile motion, we also know that the initial speed (\(u\)) at launch will equal the speed just before impact upon return. This understanding lets us connect the dots between the speed at two-thirds of the maximum height and the maximum height itself – essential to solving our textbook problem with minimal information.

Acceleration Due to Gravity

Acceleration due to gravity, denoted by the symbol \(g\) and approximately valued at \(9.8 \text{ m/s}^2\) on the surface of the Earth, is a constant force that pulls objects towards the center of the Earth. This force imparts a consistent acceleration to all objects, regardless of their mass, assuming they are in a vacuum.

When evaluating problems that involve vertical projectile motion, such as the textbook problem we're addressing, it's vital to incorporate this acceleration factor into the kinematic equations. In these scenarios, we often consider the direction upwards as positive and downwards as negative, which is why the acceleration due to gravity is taken as \-\9.8 \text{ m/s}^2\), denoting acceleration in the opposite direction to the object's initial upward motion.

For our exercise, it's the acceleration due to gravity that causes the projectile to slow down as it rises and increase in speed as it falls, effectively dictating the shape and characteristics of its trajectory – a cornerstone for calculating the maximum height reached by the object after being thrown vertically upward.