Question

Derive the formulas for time of flight, range, and maximum height in the case that an object is launched from the initial position \(\left\langle 0, y_{0}\right\rangle\) with initial velocity \(\left|\mathbf{v}_{0}\right|\langle\cos \alpha, \sin \alpha\rangle\).

Short answer

Question: Derive the formulas for time of flight, maximum height, and range for projectile motion, given an object is launched at an angle α from the ground with initial velocity |v₀| and initial position (0, y₀). Answer: The formulas for time of flight, maximum height, and range for projectile motion are: 1. Time of Flight (t_flight): \(t_{flight} = \frac{2|\mathbf{v}_{0}| \sin \alpha}{g}\) 2. Maximum Height (y_max): \(y_{max} = y_{0} + \frac{(|\mathbf{v}_{0}| \sin \alpha)^2}{2g}\) 3. Range (R): \(R = \frac{|\mathbf{v}_{0}|^2 \sin 2\alpha}{g}\)

Step-by-step solution

Determine the initial velocities for horizontal and vertical motion.

To analyze the projectile motion, we need to break up the initial velocity vector into horizontal and vertical components. The initial velocity \(\mathbf{v}_{0}\) has a magnitude \(|\mathbf{v}_{0}|\) and an angle \(\alpha\) to the horizontal. Using trigonometry, we can find the horizontal and vertical components: $$ v_{0x} = |\mathbf{v}_{0}| \cos\alpha \\ v_{0y} = |\mathbf{v}_{0}| \sin\alpha $$

Analyze vertical motion.

In vertical motion, the only force acting on the object is gravity. Therefore, the object undergoes constant acceleration with a value of \(g\) (9.81 m/s²) downwards. Thus, the equations for vertical position \(y(t)\) and vertical velocity \(v_y(t)\) at any time \(t\) are given by: $$ y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2 \\ v_y(t) = v_{0y} - gt $$

Time of Flight (ToF).

The time of flight is the time it takes for the projectile to reach the same vertical position, which is \(y_{0}\). Since we have the equation for the vertical position, we can solve for the time when \(y(t) = y_{0}\): $$ y_0 = y_0 + v_{0y}t - \frac{1}{2}gt^2 $$ Simplifying and solving for \(t\): $$ t = \frac{2v_{0y}}{g} $$ Substituting for \(v_{0y} = |\mathbf{v}_{0}| \sin\alpha\): $$ t_{flight} = \frac{2|\mathbf{v}_{0}| \sin\alpha}{g} $$

Maximum Height.

The maximum height occurs when the vertical velocity is zero, i.e., \(v_y(t) = 0\). Using the vertical velocity equation: $$ 0 = v_{0y} - gt $$ Solve for \(t\): $$ t = \frac{v_{0y}}{g} $$ Now, plug this value of \(t\) into the equation for vertical position \(y(t)\): $$ y_{max} = y_{0} + v_{0y} \frac{v_{0y}}{g} - \frac{1}{2}g(\frac{v_{0y}}{g})^2 \\ $$ Simplifying and substituting \(v_{0y} = |\mathbf{v}_{0}| \sin \alpha\): $$ y_{max} = y_{0} + \frac{(|\mathbf{v}_{0}| \sin \alpha)^{2}}{2g} $$

Analyze horizontal motion.

For horizontal motion, there is no acceleration as there are no external horizontal forces acting on the object. So, the horizontal velocity remains constant throughout the motion: $$ v_x = v_{0x} $$ The horizontal position \(x(t)\) at any time \(t\) can be given by: $$ x(t) = v_{0x}t $$

Range (R).

The range of the projectile is the horizontal distance it covers during the time of flight. Using the horizontal position equation, we want \(x(t_{flight})\): $$ R = v_{0x}t_{flight} $$ Substitute \(v_{0x} = |\mathbf{v}_{0}| \cos\alpha\) and the time of flight expression we derived earlier: $$ R = (|\mathbf{v}_{0}| \cos\alpha) (\frac{2|\mathbf{v}_{0}| \sin\alpha}{g}) \\ R = \frac{2|\mathbf{v}_{0}|^2 \sin\alpha \cos\alpha}{g} $$ By using the trigonometric identity \(2\sin\alpha \cos\alpha = \sin 2\alpha\), the range can be expressed as: $$ R = \frac{|\mathbf{v}_{0}|^2 \sin 2\alpha}{g} $$ To summarize, the formulas for time of flight, maximum height, and range for the given initial conditions are: $$ t_{flight} = \frac{2|\mathbf{v}_{0}| \sin \alpha}{g} \\ y_{max} = y_{0} + \frac{(|\mathbf{v}_{0}| \sin \alpha)^2}{2g} \\ R = \frac{|\mathbf{v}_{0}|^2 \sin 2\alpha}{g} $$

Key concepts

Time of Flight

The concept of "time of flight" in projectile motion describes how long a projectile remains in the air from the moment of launch until it returns to its original vertical position. This is crucial for understanding the duration that a projectile can stay airborne.

To calculate the time of flight, we rely on the vertical motion components. Our main element is the vertical velocity component, which, under the influence of gravity, will determine how the projectile rises and falls.

For the initial vertical velocity, denoted as \(v_{0y}\), we can use the equation:

• \(v_{0y} = |\mathbf{v}_{0}| \sin \alpha\)

Gravity acts downward, consistently decelerating and eventually reversing the upward motion. The time of flight, \(t_{flight}\), is found by setting the ending vertical position equal to the starting position and solving the equation for time:

• \(t_{flight} = \frac{2 |\mathbf{v}_{0}| \sin \alpha}{g}\)

This formula shows that the time of flight depends on both the initial launch angle and speed, staying in the air longer for higher angles or speeds.

Range

The "range" of a projectile motion refers to the horizontal distance it covers during its flight. This is an important measure when determining how far a projectile can travel before landing.

The key to finding the range is the horizontal motion of the projectile, which, unlike vertical motion, is unaffected by gravity. This makes the horizontal velocity constant throughout the flight. The horizontal component of the initial velocity is:

• \(v_{0x} = |\mathbf{v}_{0}| \cos \alpha\)

With this constant velocity, the horizontal distance, or range \(R\), is simply this velocity multiplied by the time of flight \(t_{flight}\). Therefore, the range can be calculated using the equation:

• \(R = \frac{|\mathbf{v}_{0}|^2 \sin 2\alpha}{g}\)

This clearly indicates that besides the initial velocity, the angle of projection significantly affects the range. Specifically, the optimal angle for maximum range is 45 degrees, assuming other conditions stay constant. This maximizes the \(\sin 2\alpha\) term in the equation.

Maximum Height

In projectile motion, the "maximum height" is the peak vertical position that the projectile reaches relative to its starting point. It is a crucial consideration in problems where altitude is a limiting factor or a goal.

This height is achieved when the vertical component of the projectile's velocity becomes zero momentarily before gravity starts pulling it back down. We use the initial vertical velocity again, but this time we solve for the time \(t\) when the velocity is zero:

• \(0 = v_{0y} - gt\)
• \(t = \frac{v_{0y}}{g}\)

Substitute this value into the equation for vertical position \(y(t)\), and you'll find the expression for the maximum height \(y_{max}\):

• \(y_{max} = y_{0} + \frac{(|\mathbf{v}_{0}| \sin \alpha)^2}{2g}\)

As this formula shows, not only the speed and angle of launch affect the maximum height but also any specified initial height \(y_{0}\) should be considered when calculating the total rise of the projectile.