Question

\(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) are two values of time of a projectile at the same height \(t \mathrm{t}_{1}+\mathrm{t}_{2}=\) (A) Time to reach maximum height (B) flight time for the projectile (C) \((3 / 4)\) time of the flight time. (D) \((3 / 2)\) time of the flight time.

Short answer

The relationship between \(t_1\) and \(t_2\) is given by \(t_1 + t_2 = \frac{2v_0\sin\theta}{g}\), where \(v_0\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity. This expression represents the total flight time of a projectile, making the correct answer (B) flight time for the projectile.

Step-by-step solution

Write down the equation for the height of a projectile as a function of time

The height of a projectile at a given time can be represented as: \[ h(t) = h_{0} + v_{0}\sin\theta \cdot t - \frac{1}{2}gt^2 \] where - \(h_{0}\) is the initial height (we can assume it to be 0 for simplicity), - \(v_{0}\) is the initial velocity, - \(\theta\) is the launch angle, - \(g\) is the acceleration due to gravity (\(9.81\,m/s^{2}\)), and - \(t\) is the time. With \(h_{0} = 0\), the equation becomes: \[ h(t) = v_{0}\sin\theta \cdot t - \frac{1}{2}gt^2 \] Since \(h(t_1) = h(t_2)\) we have: \[ v_{0}\sin\theta \cdot t_1 - \frac{1}{2}gt_1^2 = v_{0}\sin\theta \cdot t_2 - \frac{1}{2}gt_2^2 \]

Solve for \(t_1 + t_2\) from the equation

By solving the equation we got earlier, we can get the value of \(t_1 + t_2\). Let's rearrange the equation to get \(t_1 + t_2\) as follows: \[ v_{0}\sin\theta \cdot (t_1 - t_2) = \frac{1}{2}g(t_2^2 - t_1^2) \] \[ v_{0}\sin\theta \cdot (t_1 - t_2) = \frac{1}{2}g(t_2 + t_1)(t_2 - t_1) \] Now, dividing both sides by \((t_1 - t_2)\), we get: \[ v_{0}\sin\theta = \frac{1}{2}g(t_2 + t_1) \] Now, solving for \(t_1 +t_2\): \[ t_1 + t_2 = \frac{2v_0\sin\theta}{g} \]

Identify the relationship between \(t_1\) and \(t_2\)

From Step 2, we derived that \(t_1 + t_2 = \frac{2v_0\sin\theta}{g}\). This is the expression for the total flight time of a projectile launched at an angle \(\theta\) with an initial velocity \(v_0\). Therefore, the correct option is: (B) flight time for the projectile

Key concepts

Equation of Motion

In physics, the equation of motion is essential for understanding how an object travels through space over time. For a projectile, its motion is often described in terms of its height and horizontal distance. The equation to determine the height of a projectile is:\[ h(t) = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2 \]This equation helps us find the height at any time \(t\). Here's a breakdown:

• Initial velocity \(v_0\): The speed at which the projectile is launched.
• Launch angle \(\theta\): The angle at which the projectile is fired.
• Gravity \(g\): The gravitational pull on the projectile. On Earth, it's approximately \(9.81\,\text{m/s}^2\).

In our exercise, we looked at specific times \(t_1\) and \(t_2\) when the projectile is at the same height. By setting up an equation equating \(h(t_1)\) and \(h(t_2)\), we can explore relationships between these two instances. This involves algebraically manipulating the motion equation to gain insights about flight duration and other features.

Time of Flight

The time of flight is the total time a projectile remains in the air. It’s a crucial aspect of projectile motion as it affects how far the projectile travels horizontally. If we break down its dynamics:- When a projectile is launched, it first rises until it reaches its peak (maximum height) and then descends until it hits the ground.- The time taken to reach the peak is half of the total time of flight.In our exercise, using projectile motion concepts, it was proven that the sum of \(t_1\) and \(t_2\) equals the entire time of flight for a projectile. Solving the rearranged motion equation gave us:\[ t_1 + t_2 = \frac{2v_0\sin\theta}{g} \]This signifies how projectile's initial velocity, launch angle, and gravity determine the time of flight. Understanding this helps in predicting how long a projectile will remain airborne and where it will land.

Acceleration Due to Gravity

Gravity accelerates all objects downward on Earth. This downward force is denoted as \(g\) and has a calculated value of approximately \(9.81\,\text{m/s}^2\). It plays a vital role in projectile motion as it determines the trajectory of the projectile.When analyzing projectile motion:- Gravity decelerates the projectile as it climbs to its maximum height, causing it to momentarily stop.- Conversely, after reaching this peak height, gravity pulls it back down, accelerating its descent until it hits the ground.
Thus, the effects of gravity on a projectile are constant and ever-present throughout the projectile's path. In our equation of motion:\[ h(t) = v_0\sin\theta \cdot t - \frac{1}{2}gt^2 \]The term \(-\frac{1}{2}gt^2\) represents the impact of gravity over time, subtracting this displacement from the initial upward path. A strong understanding of gravity's role enables us to better predict, calculate, and appreciate the journey of projectiles in motion.