Question

Use \(a(t)=-32\) feet per second per second as the acceleration due to gravity. Show that the height above the ground of an object thrown upward from a point \(s_{0}\) feet above the ground with an initial velocity of \(v_{0}\) feet per second is given by the function $$ f(t)=-16 t^{2}+v_{0} t+s_{0} $$.

Short answer

The height \(f(t)\) of an object at time \(t\) thrown upwards with an initial velocity \(v_{0}\) and initial height \(s_{0}\) is given by the function \(f(t)=-16t^{2}+v_{0}t+s_{0}\). This equation is derived by sequentially integrating the expression for the acceleration due to gravity to obtain first the velocity and then the position as functions of time, using the given initial conditions.

Step-by-step solution

Establish the initial conditions and the function for acceleration

From the problem, it is given that the initial velocity \(v_{0}\) and the initial height \(s_{0}\). The acceleration due to gravity \(a(t)\) is defined to be \(a(t) = -32ft/s^{2}\). This is a constant acceleration hence does not depend on time.

Obtain velocity function by integrating the acceleration

Acceleration is the derivative of velocity with respect to time. Therefore, to obtain the velocity as a function of time, integrate the acceleration with respect to time. On integrating \(a(t)\), we get \(v(t) = -32t + C\). To find \(C\), use the initial condition \((t=0, v=v_{0})\) to get \(v_{0} = C\). Substituting \(C\) in the equation gives us the velocity function as \(v(t)=-32t+v_{0} \).

Obtain position function by integrating the velocity

Velocity is the derivative of position with respect to time. To obtain the position as a function of time, integrate the velocity function with respect to time. We get \(s(t) = -16t^{2} + v_{0}t + C'\). To find \(C'\), use the initial condition \((t=0, s=s_{0})\) to get \(s_{0}=C'\). Substituting \(C'\) in the equation gives us the position function as \(s(t) = -16t^{2} + v_{0}t + s_{0} \). That's the function we wanted to show.

Key concepts

Acceleration Due to Gravity

Explaining the concept of 'acceleration due to gravity' is essential when we discuss objects moving under the sole influence of gravity. In our context, acceleration due to gravity is a constant value that represents the rate at which an object's velocity changes per unit time due to the pull of Earth's gravity. It's typically denoted as \(g\) and has a standard approximate value of \(9.81 m/s^2\) on Earth's surface, but in the exercise, we're using \(32 ft/s^2\) due to the units being in feet.

This acceleration is always directed towards the center of the Earth and it's important to note that, in the absence of air resistance, all objects accelerate at this same rate regardless of mass. That's why our exercise uses a constant acceleration \(a(t) = -32 ft/s^2\), with the negative sign indicating that gravity is pulling the object back towards Earth.

Initial Velocity

The term 'initial velocity' refers to the velocity of an object at the start of its observation or at the initial time \(t=0\). It's often denoted as \(v_0\) and is a crucial piece of the puzzle in calculating the future position of an object in motion. In the exercise, initial velocity is what an object possesses before the influence of gravity starts to modify its motion.

Understanding Initial Velocity in the Context of the Exercise

In our scenario, the object is thrown upward with an initial velocity \(v_0\). If the velocity is positive, the object starts by moving upwards; if it's negative, it starts by moving downwards. If the initial velocity were zero, it would mean the object is just dropped rather than thrown. When we integrate the acceleration due to gravity to get the velocity function, we take into account that the object has been given this initial push (or lack thereof) in our calculations.

Position Function

The 'position function' allows us to predict where an object will be at any given time while it is in motion. It describes the location of the object, typically along a straight line path, as a function of time. For objects moving under the influence of gravity alone, the position function is a quadratic equation due to the constant acceleration of gravity.

Derivation in Our Exercise

In the exercise, we derived the position function \(s(t)\) from integrating the velocity function. The result, \(s(t) = -16t^{2} + v_0t + s_0\), is a quadratic equation representing the height of an object above the ground at any time \(t\). The term \(s_0\) represents the initial height above the ground, the term \(v_0t\) accounts for the initial velocity, and the term \(16t^2\) reflects the distance fallen under gravity's acceleration, which is squared due to the process of integration from acceleration to position. This comprehensive formula allows us to accurately chart the object's trajectory.