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Maximum height Suppose a baseball is thrown vertically upward from the ground with an initial velocity of \(v_{0} \mathrm{ft} / \mathrm{s} .\) The approximate height of the ball (in feet) above the ground after \(t\) seconds is given by \(s(t)=-16 t^{2}+v_{0} t\) a. What is the height of the ball at its highest point? b. With what velocity does the ball strike the ground?

Short Answer

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b. What is the velocity with which the ball strikes the ground in terms of the initial velocity (\(v_0\))? a. The height of the ball at its highest point is \(\frac{v_0^2}{64}\) feet. b. The velocity with which the ball strikes the ground is \(-v_0 \,\mathrm{ft/sec}\).

Step by step solution

01

Find the derivative of the height function with respect to time

To find the maximum height, we first need to find the highest point in the function which corresponds to the maximum value in the function. When the function is at its peak, the derivative with respect to time is 0. Let's calculate the derivative of \(s(t)=-16t^2+v_0t\). \(\frac{ds}{dt} = -32t + v_0\)
02

Find the time when the derivative is equal to 0

Now, let's find the time when \(\frac{ds}{dt}\) is equal to 0 to find the highest point of the ball: \(-32t + v_0 = 0\) \(t = \frac{v_0}{32}\)
03

Find the maximum height of the ball

By plugging this time into the original function, we can find the maximum height the ball reaches: \(s(\frac{v_0}{32}) = -16(\frac{v_0}{32})^2 + v_0(\frac{v_0}{32})\) \(s(\frac{v_0}{32}) = \frac{v_0^2}{64}\) Thus, the maximum height of the ball is \(\frac{v_0^2}{64}\) feet. a. The height of the ball at its highest point is \(\frac{v_0^2}{64}\) feet.
04

Find the time when the ball hits the ground

To find the velocity when the ball strikes the ground, let's first find the time when the ball hits the ground (when the height becomes 0): \(-16t^2 + v_0t = 0\) This can be factored as \(t(-16t + v_0) = 0\). Since the initial time (\(t = 0\)) corresponds to the ball leaving the ground, the other solution corresponds to the time when the ball hits the ground: \(-16t + v_0 = 0\quad\Rightarrow\quad t = \frac{v_0}{16}\)
05

Find the velocity when the ball hits the ground

Now, we can plug this time back into the derivative of the height function with respect to time (which represents the velocity) to find the velocity when the ball hits the ground: \(-32t + v_0 = -32(\frac{v_0}{16}) + v_0\) \(-32t + v_0 = -2v_0 + v_0\) So, the ball strikes the ground with a velocity of \(-v_0\) ft/s (negative sign indicates downward direction). b. The velocity with which the ball strikes the ground is \(-v_0 \,\mathrm{ft/sec}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Position Function
Understanding the derivative of a position function is crucial when analyzing motion. In the context of a projectile, like our baseball, the position function represents the height above the ground, defined by the equation \( s(t)=-16t^2+v_0t \). The derivative of this function with respect to time \( \frac{ds}{dt} \) is interpreted as the velocity of the projectile.
Taking the derivative, we obtain \( \frac{ds}{dt} = -32t + v_0 \), which represents the instantaneous rate of change of height. When this derivative equals zero, it indicates that the height has reached a peak—no more increases, no more decreases, just a moment of stillness before the projectile begins its descent. This is why setting the velocity equal to zero is the key first step in finding the maximum height of a projectile in motion.
Projectile Motion in Calculus
Projectile motion can be dissected elegantly using calculus. This type of motion describes a projectile like a baseball being thrown into the air and experiencing the Earth's gravitational pull. The vertical motion is typically represented by a quadratic function, for instance, \( s(t)=-16t^2+v_0t \), where \( s(t) \) is height, \( t \) is time, and \( v_0 \) is the initial velocity.
Analysing this motion involves computing the maximum height and the velocity when the projectile returns to the ground. Calculus tools, such as finding the derivative to determine the peak height (maximum value of the function), or solving for zeroes to determine when the object lands, are applied for these purposes. The constants integrated into the function, like the -16 in our equation, reflect real-world forces like gravity, making the analysis grounded in physical reality.
Optimization Problem in Calculus
Optimization problems in calculus often deal with finding maximum or minimum values of functions. In our problem, the optimization aspect comes into play by seeking the maximum height of the baseball. The process involves finding the derivative of the function describing the height of the ball over time, setting it equal to zero, and solving for the time at which the height is maximized. In this case, the quadratic function has a single maximum, as it models the ascent and descent of the ball in a parabolic trajectory.
After identifying the critical point by setting the derivative equal to zero, the time obtained is substituted back into the original position function to determine the maximum height. The key is understanding that at the maximum height, the derivative (velocity) is zero because the upward motion ceases momentarily before descent commences. This fundamental concept underlies not only projectile problems but a wide variety of optimization scenarios in calculus.
Kinematics in Calculus
Kinematics in calculus refers to the study of motion without considering the forces that cause it. It involves analyzing the relationships between position, velocity, and acceleration over time. These relationships are often expressed through functions and their derivatives. For a projectile, the initial position\( s(0) \) is typically the launching point, the first derivative of the position \( s'(t) \) expresses the velocity, and the second derivative \( s''(t) \) corresponds to the acceleration.
In the exemplary case of the thrown baseball, the acceleration due to gravity is constant, represented in the equation as \( -16t^2 \). However, in calculus-based kinematics, you can also encounter more complex scenarios where acceleration isn't constant, requiring integration to uncover the velocity and position functions. This field is vital in fields ranging from engineering to video game programming, anytime there's motion to be modeled and predicted.

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