Question

Free-Falling Object In Exercises 103 and 104 , use the position function \(s(t)=-4.9 t^{2}+200\) , which gives the height (in meters) of an object that has fallen for \(t\) seconds from a height of 200 meters. The velocity at time \(t=a\) seconds is given by \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\) At what velocity will the object impact the ground?

Short answer

The velocity of the object at the moment of impact can be found by first determining the time of impact given by solving \(s(t)=0\) for \(t\) in the position function \(s(t)=-4.9 t^{2}+200\), and then plugging this time into the limit equation \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\), which works out to the velocity function. Simplify this function and substitute the value of \(t\) into it to obtain the object's impact velocity.

Step-by-step solution

Determine time of impact

To calculate the time when the object makes impact with the ground, when the height 's' is zero, insert \(s(t)=0\) into the function \(s(t)=-4.9 t^{2}+200\) and find the value for \(t\).

Compute the velocity

Substitute the time 't' calculated in Step 1 into the limit equation \(\lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t}\) to find the velocity of the object at the moment of impact.

Simplify and Evaluate

To simplify, substitute 's(a)' with \(-4.9 a^{2}+200\) and 's(t)' with \(-4.9 t^{2}+200\). This will give you the equation for the difference quotient, which will simplify to the derivative of the position function, yielding the velocity function. Plug in the given \(t\), the time when the object hits the ground, into this velocity function to find the velocity at the moment of impact.

Key concepts

Position Function

Understanding the position function is critical when studying motion in physics and calculus. The position function describes how the position of an object changes over time. In the context of free fall, as in our exercise, the position function takes the form of a quadratic equation. Here, the function is given by s(t) = -4.9t^2 + 200, where s(t) denotes the height of the object above the ground in meters, after t seconds of falling from an initial height of 200 meters.

The negative coefficient of the t^2 term indicates that the object is moving downward due to gravity accelerating it in a negative direction (towards the Earth). Gravity is traditionally taken as 9.8 m/s^2 for free-falling objects near Earth's surface, and thus -4.9 is half of that value, aligning with the kinematic equations of motion under constant acceleration.

This quadratic form is very standard in physics for free-fall scenarios because it incorporates both the effects of gravity and the initial conditions—in this case, an initial height of 200 meters, with no initial velocity, as implied by the lack of a linear t term.

Velocity in Free Fall

Grasping the Concept of Velocity

The concept of 'velocity in free fall' refers to how fast an object is moving and in which direction during its descent. In a free-fall situation, gravity is the only force acting on the object, and it causes the object to accelerate downwards. The calculation of velocity at any point in time is vital to understanding the dynamics of the motion.

In our exercise, we look at the velocity of an object as it impacts the ground. Velocity is a vector quantity, meaning it has both magnitude and direction; here, we are mainly concerned with the magnitude, sometimes referred to as speed, and its direction is downward. To calculate this, we use the concept of a limit from calculus, which enables us to find the instantaneous rate of change of position (or velocity) as time approaches a specific value.

Limits and Derivatives

The concepts of limits and derivatives are fundamental in calculus and are particularly important for analyzing motion. The derivative of a function at a given point is the limit of the difference quotient as the interval approaches zero. In simpler terms, it represents the instant rate of change of the function, which, in the context of motion, corresponds to velocity.

In our free-fall problem, the derivative of the position function s(t) gives us the velocity function v(t), which indicates how fast the object is moving at any moment in time. To find this, we take the limit of the difference quotient as t approaches a specific time point a. This is represented by the limit formula provided in our exercise: \( \lim _{t \rightarrow a} \frac{s(a)-s(t)}{a-t} \).

Calculating Instantaneous Velocity

By simplifying the difference quotient, we actually calculate the derivative of s(t) with respect to time t, which in turn gives us the object's instantaneous velocity as it impacts the ground. Derivatives in this context help us to understand not just the speed of the object at impact, but also how its velocity changes throughout its fall, providing a more comprehensive picture of its motion.