S(t) = s0 + v0t + 1/2gt2
where v0 is its initial velocity and g is the constant of acceleration due to gravity. Suppose we take the measurements shown in Table 7.6.
(a) Find the least squares approximating quadratic for these data.
(h) Estimate the height at which the object was released (in m), its initial velocity (in m/s), and its acceleration due to gravity (in m/s2).
(c) Approximately when will the object hit the ground?
The Correct Answer and Explanation is :
Let’s break this down into three parts as you’ve outlined:
Part (a): Least Squares Approximating Quadratic
We are provided with the equation of motion for an object in free fall: S(t)=s0+v0t+12gt2S(t) = s_0 + v_0 t + \frac{1}{2} g t^2
Where:
- S(t)S(t) is the position at time tt.
- s0s_0 is the initial height.
- v0v_0 is the initial velocity.
- gg is the acceleration due to gravity (approximately 9.8 m/s², but we will estimate it from the data).
To find the least squares approximating quadratic for the data, we would need the actual measurements (i.e., the values for time and height). In general, you would use a method of least squares to fit a quadratic function of the form S(t)=a+bt+ct2S(t) = a + bt + ct^2 to the data points. This can be done by solving a system of equations derived from minimizing the sum of the squared differences between the observed data and the quadratic model.
We can assume that the system would look something like: Minimize∑i=1n(S(ti)−(a+bti+cti2))2\text{Minimize} \sum_{i=1}^{n} (S(t_i) – (a + b t_i + c t_i^2))^2
Once the least squares coefficients aa, bb, and cc are determined, the quadratic equation will describe the object’s motion.
Part (b): Estimate the Height, Initial Velocity, and Acceleration
We want to estimate the following parameters:
- Initial height s0s_0
- Initial velocity v0v_0
- Acceleration due to gravity gg
We can obtain these estimates from the coefficients of the least squares fit. From the general form of the equation: S(t)=s0+v0t+12gt2S(t) = s_0 + v_0 t + \frac{1}{2} g t^2
- a=s0a = s_0, so the initial height is the constant term of the quadratic fit.
- b=v0b = v_0, so the initial velocity is the coefficient of tt.
- c=12gc = \frac{1}{2} g, so the acceleration due to gravity gg can be calculated as g=2cg = 2c.
Part (c): Time When the Object Hits the Ground
The object hits the ground when S(t)=0S(t) = 0, meaning the height becomes zero. The equation to solve for the time when this occurs is: 0=s0+v0t+12gt20 = s_0 + v_0 t + \frac{1}{2} g t^2
This is a quadratic equation in tt, and it can be solved using the quadratic formula: t=−v0±v02−2gs0gt = \frac{-v_0 \pm \sqrt{v_0^2 – 2g s_0}}{g}
This will give two possible values for tt. Since time cannot be negative, we take the positive root as the solution for when the object hits the ground.
Explanation:
To summarize, you would first perform a least squares fitting of a quadratic equation to the data. Once the coefficients are found, they would provide estimates for the initial height s0s_0, initial velocity v0v_0, and the acceleration due to gravity gg. With these parameters, we then solve the quadratic equation for when the object reaches the ground, that is, when S(t)=0S(t) = 0. The solution provides the approximate time of impact.