Question

Parabolic trajectories Show that the two-dimensional trajectory \(x(t)=u_{0} t+x_{0} \quad\) and \(\quad y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0}, \quad\) for \(0 \leq t \leq T\) of an object moving in a gravitational field is a segment of a parabola for some value of \(T > 0 .\) Find \(T\) such that \(y(T)=0\)

Short answer

Question: Show that the trajectory of an object in a gravitational field with given parametric equations follows a parabolic path and find the time when the object reaches y=0. Answer: The object's position can be represented with the quadratic form \(y = Ax^2 + Bx + C\), which confirms that the trajectory is a segment of a parabola. The time when \(y(T) = 0\) is given by \(T = \frac{2v_0 + \sqrt{(2v_0)^2 + 8gy_0}}{2g}\).

Step-by-step solution

Understand the given equations

The given equations are parametric equations for the position of an object in a gravitational field, given by: \(x(t) = u_0 t + x_0\) for the horizontal position, \(y(t) = -\frac{g t^2}{2} + v_0 t + y_0\) for the vertical position, where \(u_0\) and \(v_0\) are the initial velocity components, \(x_0\) and \(y_0\) are the initial position, and \(g\) is the acceleration due to gravity.

Solve for t from the x(t) equation

In order to eliminate \(t\) from the equations, we'll first solve for \(t\) in the \(x(t)\) equation: $$t=\frac{x-x_0}{u_0}$$

Substitute t in the y(t) equation

Now substitute the expression for \(t\) in the \(y(t)\) equation: $$y=-\frac{g}{2} \left(\frac{x-x_0}{u_0}\right)^2 + v_0\left(\frac{x-x_0}{u_0}\right) + y_0$$

Identify the parabolic form

The equation we have now represents a parabola. It's in the quadratic form of \(y = Ax^2 + Bx + C\), with \(A = -\frac{g}{2u_0^2}\), \(B = \frac{v_0}{u_0}\), and \(C = y_0-\frac{gx_0^2}{2u_0^2}\). This confirms that the trajectory of the object is indeed a segment of a parabola.

Find the time T when y(T) = 0

Now, we are asked to find the time \(T\) when \(y(T) = 0\). Substitute \(y = 0\) in the \(y(t)\) equation and solve for \(t\): $$0 = -\frac{gT^2}{2} + v_0T + y_0$$ $$\Rightarrow gT^2 - 2v_0T - 2y_0 = 0$$ This is a quadratic equation in \(T\). Using the quadratic formula, we find: $$T = \frac{2v_0 \pm \sqrt{(2v_0)^2 - 4(-2y_0)g}}{2g}$$ We can discard the negative solution as time must be positive. Therefore, the time \(T\) when \(y(T) = 0\) is given by: $$T = \frac{2v_0 + \sqrt{(2v_0)^2 + 8gy_0}}{2g}$$

Key concepts

Parametric Equations

Parametric equations provide a convenient way to represent curves in mathematics, where each point on the curve is defined by functions of one or more parameters. In the context of parabolic trajectories, parametric equations express the position of an object as a function of time.

For example, the equations \(x(t) = u_0 t + x_0\) and \(y(t) = -\frac{1}{2}gt^2 + v_0 t + y_0\) use time (\(t\)), to define the horizontal position (\(x(t)\)) and the vertical position (\(y(t)\)) of an object in motion. The parameters \(u_0\) and \(v_0\) represent the initial velocity components, while \(x_0\) and \(y_0\) denote the initial positions.

To visualize parametric equations, one can plot the calculated positions (\(x(t)\) and \(y(t)\)) at different moments in time to trace the path of the object, which in this case is a segment of a parabola.

Gravitational Field Motion

An object moving under the influence of gravity alone, without any air resistance or other forces, follows a parabolic trajectory. This motion is a classic example of a physical phenomenon called projectile motion.

The gravitational field exerts a constant acceleration \(g\), directed downwards, which affects the vertical position \(y(t)\) of the object. The acceleration causes the vertical velocity to decrease over time until reaching the peak of the trajectory, after which it increases in the negative direction as the object falls back towards the ground.

Gravitational field motion is perfectly symmetrical in a vacuum, meaning the path taken upwards until the peak is mirrored in the downward path. Understanding this concept helps to grasp why the object's path forms a parabola and to predict its behavior at any point in time, providing the initial conditions are known.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra for finding the solutions to quadratic equations. A quadratic equation is any equation that can be written in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.

The formula to find the roots of the quadratic equation is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. This equation provides two solutions, corresponding to the points where a parabola crosses the x-axis. In the context of parabolic trajectories, when the question arises at what time \(T\) the object will hit the ground (assuming it started above ground level), the quadratic formula can be applied to the second parametric equation to find the moment in time when \(y(T)\) will be zero.

One must carefully consider which of the two solutions from the quadratic formula is physically meaningful, as negative time would be non-physical in this context. Therefore, only the positive solution in time corresponds to the actual event when the object hits the ground.