Question

Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)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: Determine the time T at which an object thrown upwards returns to the same height it started from, given the equations of motion $$x(t)=u_{0} t+x_{0}$$ and $$y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0}$$. Answer: The time T at which the object returns to the same height it started from is given by $$T=\frac{v_{0}+\sqrt{v_{0}^2+2gy_{0}}}{g}$$.

Step-by-step solution

Identify the equations of motion

In this problem, the equations of motion for the object in the two dimensions are given by: $$x(t)=u_{0} t+x_{0}$$ $$ y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0}$$

Analyze the given trajectory

We can see that the x(t) equation is a linear function, while the y(t) equation is a quadratic function. Quadratic functions represent parabolas, so we can infer that the trajectory is a parabola segment in 2D space.

Find the value of T for y(T)=0

To find the value of T such that \(y(T)=0\), we need to solve the y(t) equation for t when y=0: $$ 0=-\frac{g T^{2}}{2}+v_{0} T+y_{0}$$

Rearrange and solve for T

Rearrange the equation to make it a quadratic equation and then solve for T: $$\frac{g T^{2}}{2}-v_{0} T-y_{0}= 0$$ Quadratic equations have the form \(aT^2 + bT + c = 0\), where in this case \(a=\frac{g}{2}\), \(b=-v_{0}\), and \(c=-y_{0}\). To find T, we can use the quadratic formula: $$T=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Plug in the values for a, b, and c and solve for T: $$T=\frac{v_{0}\pm\sqrt{(-v_{0})^2-4(\frac{g}{2})(-y_{0})}}{g}$$ Since the object returns to the ground, we consider only the positive root for a valid result: $$T=\frac{v_{0}+\sqrt{v_{0}^2+2gy_{0}}}{g}$$ Now we have found the value of T such that \(y(T)=0\).

Key concepts

Quadratic Equation

Quadratic equations are essential in understanding the motion of objects in certain physical contexts, like a projectile in a gravitational field. A quadratic equation typically takes the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our exercise, the quadratic equation describes the vertical component \( y(t) \) of the object's motion:

• \( y(t) = -\frac{g t^2}{2} + v_0 t + y_0 \)

Here, the object is moving under gravity. The term \( -\frac{g t^2}{2} \) represents the gravitational effect, and \( v_0 t \) represents the initial velocity affecting the vertical position. This quadratic formula helps us find important values, like \( T \), which is the time when the object hits the ground (i.e., \( y(T) = 0 \)).
We solve the quadratic equation using the quadratic formula:\[T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where:

• \( a = \frac{g}{2} \)
• \( b = -v_0 \)
• \( c = -y_0 \)

The positive root gives the exact time \( T \) when the object returns to its starting vertical height. Understanding the quadratic nature of the equation is key to predicting the projectile's future position.

Parabola

A parabola is the graphical representation of a quadratic equation. It is a U-shaped curve on a graph. When we analyze the equation \( y(t) = -\frac{g t^2}{2} + v_0 t + y_0 \), we quickly find it represents a parabola. This is due to the squared term \( -\frac{g t^2}{2} \), controlling how the path curves.
In our problem, the object's path is part of a parabola due to gravity's influence. The gravity creates the characteristic arch shape. The parabolic trajectory tells us a lot about the motion:

• The vertex of the parabola indicates the highest point of the trajectory.
• The focus and directrix provide deeper insights into how steep or wide the curve is.
• The axis of symmetry, a vertical line running through the vertex, divides the parabola into two symmetrical halves.

In projectile motion, we're typically interested in the portion of the parabola that starts from launch and lands back to the ground. By transforming the quadratic expression, we explore these aspects and how they influence the object's trajectory. Understanding these elements helps us find specific moments, like when \( y(T)=0 \), where the object hits the ground again.

Gravitational Field

The gravitational field is a force that acts on objects with mass. In the context of projectile motion, this field pulls the object towards the earth, impacting its trajectory. This force modifies the parabolic path described by the quadratic equation for \( y(t) \).

• Gravity acts downward with a magnitude represented by \( g \), the acceleration due to Earth's gravity, approximately \( 9.8 \text{ m/s}^2 \).
• In the equation \( y(t) = -\frac{g t^2}{2} + v_0 t + y_0 \), \( g \) is crucial as it affects the curvature of the path, making the trajectory a parabola.
• The term \( -\frac{g t^2}{2} \) specifically accounts for the gravitational pull on the object as time progresses.

Gravitational fields always exert a constant acceleration in projectile motion. They determine both the rate of descent and the symmetry of motion: both ascent and descent mirror each other under ideal conditions.
Understanding gravity's role helps explain characteristics like the maximum height, time of flight (\( T \)), and range of the projectile. Thus, when analyzing projectile motion problems, knowing how a gravitational field exerts its influence is key to solving equations and predicting behavior accurately.