Question

A particle is thrown in upward direction with Velocity \(\mathrm{V}_{0}\). It passes through a point \(\mathrm{p}\) of height \(\mathrm{h}\) at time \(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) so \(\mathrm{t}_{1}+\mathrm{t}_{2}=\ldots\) (A) \(\left(\mathrm{v}_{0} / \mathrm{g}\right)\) (B) \(\left[\left(2 \mathrm{v}_{0}\right) / \mathrm{g}\right]\) (C) \((2 \mathrm{~h} / \mathrm{g})\) (D) \((\mathrm{h} / 2 \mathrm{~g})\)

Short answer

The short answer is: \(t_1 + t_2 = \frac{2V_{0}}{g}\)

Step-by-step solution

Identify the variables

Let's identify all the known and unknown variables. - Initial velocity (Vo) - Height (h) - Time taken to reach height h (t1 and t2) - Acceleration due to gravity (g)

Use the equations of motion

Since the motion is under gravity, we use the equations of motion for an object under gravity. The equation for height (y) in terms of time (t) is given by: \(y = V_{0}t - \frac{1}{2}gt^2\) At point P, the height is h. So, we need to solve for t: \(h = V_{0}t - \frac{1}{2}gt^2\)

Solve the equation for t

In order to solve for t, rewrite the equation as a quadratic equation: \(\frac{1}{2}gt^2 - V_{0}t + h = 0\) Now we can use the quadratic formula to solve for t: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In our case, a = 1/2 * g, b = -Vo, and c = h.

Calculate the values of t

Plugging these values into the quadratic formula, we get: \(t_1 = \frac{Vo + \sqrt{V_{0}^2 - 2gh}}{g}\) and \(t_2 = \frac{Vo - \sqrt{V_{0}^2 - 2gh}}{g}\) We have to find the sum of these times t1 + t2.

Calculate t1 + t2

Add t1 and t2: \(t_1 + t_2 = \frac{Vo + \sqrt{V_{0}^2 - 2gh}}{g} + \frac{Vo - \sqrt{V_{0}^2 - 2gh}}{g}\) Simplify the expression: \(t_1 + t_2 = \frac{2Vo}{g}\)

Choose the correct option

Now that we have found the value of t1 + t2, we can match it with the given answer choices. Answer: (B) \(\left[\left(2 \mathrm{v}_{0}\right) / \mathrm{g}\right]\)

Key concepts

Equations of Motion

Projectile motion is a beautiful demonstration of the laws of physics, specifically the equations of motion. These equations help us understand how objects move through space over time. When a particle is thrown upwards with an initial velocity, its motion is influenced by gravity, which pulls it back to the ground. The primary equation we use is \[ y = V_{0}t - \frac{1}{2}gt^2 \]where:

• \( y \) is the height of the projectile at time \( t \)
• \( V_{0} \) is the initial velocity
• \( g \) is the acceleration due to gravity, typically \( 9.8 \, m/s^2 \)

These equations help us predict how high the object will go and how long it will stay in the air.
This simple model provides the foundation for more complex analyses in physics, like the motion of satellites and rockets.

Quadratic Equation

To solve problems in projectile motion, especially when identifying time variables, we use quadratic equations. The equation of motion we derived earlier:\[ \frac{1}{2}gt^2 - V_{0}t + h = 0 \]is a quadratic equation in standard form: \[ at^2 + bt + c = 0 \]where

• \( a = \frac{1}{2}g \)
• \( b = -V_{0} \)
• \( c = h \)

Solutions to this quadratic equation can be found using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula gives us two possible values of \( t \), which represent the times when the projectile is at the same height on its upward and downward paths. Quadratic equations are pivotal in solving many real-world problems, bridging mathematics and physics.

Time of Flight

Time of flight refers to the total time a projectile is in motion. For an object thrown upward, it reaches a peak height and then descends back down. The problem of calculating time of flight often involves determining these individual times when the projectile reaches a particular height.
The sum \( t_{1} + t_{2} \) gives us useful insight into this motion, derived as:\[ t_{1} + t_{2} = \frac{2V_{0}}{g} \]This expression shows that the total time a projectile spends in the air is twice the quotient of initial velocity over acceleration due to gravity. Understanding these time intervals allows physicists and engineers to design paths and predict where projectiles will land, valuable in fields from sports to aerospace.

Initial Velocity

The initial velocity \( V_{0} \) is the speed at which a projectile is launched. It plays a crucial role in determining the projectile's trajectory.
Higher initial velocities mean higher peaks and longer distances covered by the projectile. The initial speed greatly influences several factors:

• The maximum height reached
• The time of flight
• The horizontal range

In our given problem, this initial velocity helps determine the times \( t_{1} \) and \( t_{2} \) and, ultimately, their sum. By impacting these times, initial velocity affects the overall path and behavior of the projectile. Athletes, engineers, and scientists must understand these dynamics to predict motion outcomes accurately.