Question

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

Short answer

Question: Derive the formulas for the time of flight, range, and maximum height for a projectile launched at an initial position of (0, y0) above the ground and an initial velocity of |v0|⟨cosα, sinα⟩. Answer: The formulas for the time of flight (T), range (R), and maximum height (H) are as follows: 1. Time of Flight (T): T = solution of 0 = y0 + v0*sinα*T - (1/2)*g*T² 2. Range (R): R = v0*cosα * T 3. Maximum Height (H): H = y0 + v0*sinα*tH - (1/2)*g*tH², where tH = v0*sinα / g

Step-by-step solution

Identify known factors and variables

We are given the initial position \((0, y_0)\), the initial velocity \(\left|\mathbf{v}_{0}\right|\langle\cos \alpha, \sin \alpha\rangle\), and the angle \(\alpha\) between the ground and the initial velocity. It's important to note that the motion of the object is under the influence of gravity, and it follows a parabolic path.

Time of Flight (T)

The time of flight, denoted as \(\displaystyle T\), is the time the projectile is in the air. To calculate \(\displaystyle T\), we need to find when the projectile lands on the ground again, that is, when its vertical position \(\displaystyle y\) is zero. The equation of motion for the vertical component can be expressed as: \(\displaystyle y=y_{0}+v_{0}\sin \alpha t-\dfrac{1}{2} g t^{2}\) Setting \(\displaystyle y=0\) and solving for \(\displaystyle t\), we get: \(\displaystyle 0=y_{0}+v_{0}\sin \alpha T-\dfrac{1}{2} g T^{2}\) Where \(\displaystyle t=T\), the time of flight.

Range (R)

The range, denoted by \(\displaystyle R\), is the horizontal distance the projectile has traveled by the time it lands back on the ground. To find \(\displaystyle R\), we can use the time of flight \(\displaystyle T\) previously calculated. Since the horizontal component of velocity remains constant due to no horizontal forces acting on the projectile, the range can be calculated using: \(\displaystyle R=v_{0}\cos \alpha \cdot T\) Where \(\displaystyle T\) is the time of flight.

Maximum Height (H)

The maximum height, denoted by \(\displaystyle H\), is the highest point the projectile reaches on its parabolic trajectory. To find \(\displaystyle H\), we must find the time at which the vertical component of the velocity (\(\displaystyle v_{y}\)) becomes zero, denoted by \(\displaystyle t_{H}\). The equation for \(\displaystyle v_{y}\) can be expressed as: \(\displaystyle v_{y}=v_{0}\sin \alpha -gt_{H}\) Setting \(\displaystyle v_{y}=0\), and solving for \(\displaystyle t_{H}\), we get: \(\displaystyle t_{H}=\dfrac{v_{0}\sin \alpha }{g}\) Now, we can use \(\displaystyle t_{H}\) to find the maximum height \(\displaystyle H\) by substituting \(\displaystyle t_{H}\) into the vertical position equation: \(\displaystyle H=y_{0}+v_{0}\sin \alpha t_{H}-\dfrac{1}{2} g t_{H}^{2}\) By substituting the values and solving, we find the formula for the maximum height \(\displaystyle H\). In conclusion, the derived formulas for the time of flight, range, and maximum height are: \(\displaystyle T=\text{solution of }0=y_{0}+v_{0}\sin \alpha T-\dfrac{1}{2} g T^{2}\) \(\displaystyle R=v_{0}\cos \alpha \cdot T\) \(\displaystyle H=y_{0}+v_{0}\sin \alpha t_{H}-\dfrac{1}{2} g t_{H}^{2}\)

Key concepts

Time of Flight

The time of flight of a projectile is the total time it remains in the air. Understanding this concept is crucial for predicting when the projectile will land back on the ground. The time of flight is mainly determined by the initial velocity, the angle of launch, and the initial height from which the projectile is launched.

Here’s how we calculate it:

• Using the equation of motion for the vertical component: \[ y = y_0 + v_0 \sin \alpha \cdot t - \frac{1}{2} g t^{2} \]
• Set the vertical position \( y \) to zero because you're calculating when it lands: \[ 0 = y_0 + v_0 \sin \alpha \cdot T - \frac{1}{2} g T^{2} \]
• Solve this equation for \( T \), which represents the time of flight.

This means you're determining at what point in time the path of the projectile completes its journey, hitting the ground again. Solving this quadratic equation will give you two values for \( T \): one is the initial time when it is launched, and the second is the time when it lands.

Range of a Projectile

The range of a projectile refers to the total horizontal distance it travels before landing. It's an essential factor when you want to calculate how far a projectile will go. Importantly, the range is influenced by the initial speed, launch angle, and height from which the projectile is thrown.

Here's a breakdown of how to find the range:

• You already know the time of flight \( T \) from the calculation of the vertical trajectory.
• Use the horizontal component of velocity, which remains constant: \[ R = v_0 \cos \alpha \cdot T \]

The product of the initial horizontal velocity and the time of flight gives you the range. This indicates how important both the launch angle and the initial velocity are for determining the reach of the projectile. A greater velocity or an angle can significantly extend how far it travels.

Maximum Height

Achieving maximum height is seeing a projectile reach the peak of its arc. This peak is where the vertical velocity momentarily reaches zero. The maximum height reflects the highest vertical point from launch and is key for understanding how much vertical displacement occurs.

Calculating maximum height involves these steps:

• Determine when the vertical speed is zero, using:\[ v_{y} = v_0 \sin \alpha - gt_{H} \]
• Solve for \( t_{H} \), the time at peak height:\[ t_{H} = \frac{v_0 \sin \alpha}{g} \]
• Then substitute \( t_{H} \) into the vertical position formula:\[ H = y_0 + v_0 \sin \alpha \cdot t_{H} - \frac{1}{2} g t_{H}^{2} \]

This calculation tells you how high the projectile will reach relative to the launch height. Understanding maximum height is vital for scenarios requiring the projectile to clear obstacles or reach certain altitudes.