Question

The height of a ball \(t\) seconds after it is thrown straight up is \(-^{1} A g t^{2}+401\) (where \(g\) is the acceleration due to gravity a. If \(g=32\) (as on the earth), when does the ball reach a maximum height? What is that height? b. If \(g=5.5\) (as on the moon), when does the ball reach a maximum height and what is that height? Can you explain the reasons for the difference between this answer and the answer for part (a)? c. In general, develop an expression for the change in maximum height for a unit change in \(g .\) Explain why this value depends implicitly on the value of \(g\) itself.

Short answer

Question: Explain why the maximum height of the ball does not depend on the value of 'g', the acceleration due to gravity. Answer: The maximum height of the ball does not depend on the value of 'g' because the coefficient 'B', which is part of the vertex formula, is 0. As a result, changing the value of 'g' has no effect on the time it takes for the ball to reach the maximum height and the maximum height itself. Furthermore, the derivative of the height equation with respect to 'g' is 0 when evaluated at the maximum height, indicating that the maximum height does not change as 'g' changes.

Step-by-step solution

Part a: Maximum height and time on Earth

Given the equation: \(h(t)=-\frac{1}{2}gt^2+401\) Let \(g=32\) 1. Find the vertex of the parabola The vertex of a parabola represented by the equation \(h(t)=At^2+Bt+C\) is given by the formula: \(t_v= -\frac{B}{2A}\) In our case, A = \(-\frac{1}{2}g\) and B = 0 So, \(t_v= -\frac{0}{2(-\frac{1}{2}(32))} = 0\) 2. Find the maximum height Plug in \(t_v=0\) into the equation to find the maximum height, \(h(0) = -\frac{1}{2}(32)(0)^2+401 = 401\) ft The ball reaches its maximum height of 401 ft in 0 seconds on Earth.

Part b: Maximum height and time on the Moon

Let \(g=5.5\) 1. Find the vertex of the parabola Using the same vertex formula, we have \(t_v =-\frac{0}{2(-\frac{1}{2}(5.5))} = 0\) 2. Find the maximum height Plug in \(t_v=0\) into the equation to find the height, \(h(0) = -\frac{1}{2}(5.5)(0)^2+401 = 401\) ft The ball reaches its maximum height of 401 ft in 0 seconds on the Moon. The answers for parts (a) and (b) are the same because the equation does not depend on the value of 'g' since the coefficient B is 0. Therefore, changing the value of 'g' has no effect on the time it takes to reach the maximum height and the maximum height itself.

Part c: Change in maximum height per unit change in 'g'

1. Find the derivative of the height equation with respect to 'g' \( \frac{dh}{dg} = \frac{d(-\frac{1}{2}gt^2+401)}{dg} = -\frac{1}{2}t^2\) 2. Evaluate the derivative at a given 'g' value To find the change in maximum height per unit change in 'g' at a specific 'g' value, plug in the value of 't_v=0$ (since that's when the maximum height occurs): \( \left.\frac{dh}{dg}\right|_{t_v=0} = -\frac{1}{2}(0)^2 = 0\) The change in maximum height per unit change in 'g' is 0. This means that the maximum height does not depend on the value of 'g' and is constant, as we observed in parts (a) and (b).

Key concepts

Parabolic Motion

Parabolic motion, also referred to as projectile motion, describes the movement of an object thrown or projected into the air that follows a curved path under the influence of gravity. This path can be described by a quadratic equation in the form of \( h(t) = At^2 + Bt + C \), where \( h(t) \) is the height of the object at time \( t \).

The motion is symmetrical, with the object ascending and descending in a similar manner on either side of the peak. To efficiently compute the attributes of parabolic motion, understanding the initial velocity, launch angle, and acceleration due to gravity is crucial. The principles of parabolic motion are not only critical for textbook exercises but are also applied in real-world scenarios such as ballistics, sports, and even spaceflight.

In the given exercise, we examined the special case where the ball is thrown straight up. This implies that there is no horizontal movement and hence no \( Bt \) term because the initial horizontal velocity is zero. Therefore, the ball's motion becomes a simplified one-dimensional problem where the only forces acting are the initial upward velocity and the constant downward acceleration due to gravity.

Acceleration Due to Gravity

Acceleration due to gravity, denoted \( g \), is a constant that measures the force of gravity acting on an object near the surface of a celestial body, such as Earth or the Moon. On Earth, \( g \) is approximately \( 9.8 \text{ m/s}^2 \), while on the Moon, it is about \( 1.6 \text{ m/s}^2 \), which is about one-sixth of Earth's value. This gravitational acceleration pulls objects towards the center of the celestial body and is a critical factor in calculating the time and distance an object will travel when in free fall.

The value of \( g \) is crucial when calculating parabolic motion because it directly influences the curvature and the time it takes for an object to reach its maximum height, as well as its descent back to the initial height. In our exercise, we see an oversimplified case where \( g \) appears to not affect the height due to an initial condition where time \( t \) is zero, meaning the object has not experienced any motion yet.

Vertex of a Parabola

The vertex of a parabola is the highest or lowest point on its graph, depending on whether it opens upwards or downwards, respectively. In projectile motion, the vertex represents the maximum height that the object reaches when thrown up into the air. Mathematically, the vertex \((t_v, h(t_v))\) can be found using the formula \( t_v=-\frac{B}{2A} \), where A and B are the coefficients from the quadratic equation \( h(t)=At^2+Bt+C \).

Since the parabola is symmetrical, the time to reach the vertex is half the total time the object is in the air. For the case presented in the exercise, the coefficient \( B \) is 0, which simplifies our vertex calculation, rendering the time to reach the vertex as zero. However, it is important to understand that in real-life scenarios, the vertex not only tells us about the maximum height but also indicates when this event occurs during the motion. This insight is essential for predicting the behavior of objects in flight and for various applications requiring precise timing of peak altitude, such as optimizing the flight path of a rocket or timing the release of a parachute.