Problem 53 - Master Resource Book in JEE Main Physics

Question

A body projected vertically upwards crosses a point twice in it journey at a height \(h\) first after \(t_{1}\) and \(t_{2}\) second. Maximum height reached by the body is [EAMCET 2005] (a) \(2 g\left(t_{1}+t_{2}\right)\) (b) \(\frac{g}{4}\left(t_{1}+t_{2}\right)^{2}\) (c) \(\frac{g}{4}\left(t_{1} t_{2}\right)\) (d) \(2 g\left(\frac{t_{1}+t_{2}}{4}\right)^{2}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(\frac{g}{4}(t_1+t_2)^{2}\).

Step 1: Understanding the Problem

We need to find the maximum height reached by a body projected vertically upwards. It crosses a certain point at a height \(h\) two times during its journey—at times \(t_1\) and \(t_2\). To solve this, we'll determine a relationship involving these times and the maximum height.

Step 2: Recall the Vertical Motion Equation

The equation of motion for vertical projection is \(h = ut - \frac{1}{2}gt^2\), where \(h\) is the height, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time.

Step 3: Apply Point Crossing Conditions

Since the body crosses the same height \(h\) twice at \(t_1\) and \(t_2\), set the height equations equal: \(h = ut_1 - \frac{1}{2}gt_1^2 = ut_2 - \frac{1}{2}gt_2^2\). This implies the body returns to the same height at these two times due to symmetrical projectile motion.

Step 4: Use the Symmetrical Property

For projectile motion, the time of ascent equals the time of descent. Therefore, \(t_1 + t_2\) is the total time to go up to the highest point and come back down to \(h\), implying the time to reach the max height is \(\frac{t_1 + t_2}{2}\).

Step 5: Determine the Maximum Height Expression

The body reaches maximum height at time \(\frac{t_1 + t_2}{2}\). We use this to find max height minding that speed at top is zero: \[v = u - g\left(\frac{t_1 + t_2}{2}\right) = 0\] implies \[u = g\left(\frac{t_1 + t_2}{2}\right)\].Then, \[\text{Max Height} = \frac{u^2}{2g} = \frac{\left(g\left(\frac{t_1 + t_2}{2}\right)\right)^2}{2g} = \frac{g(t_1 + t_2)^2}{8}\].

Step 6: Match the Option

Among the options given, option (b) matches \(\frac{g}{4}(t_1 + t_2)^2 \) which can be simplified as \(\frac{g(t_1 + t_2)^2}{4} \times \frac{1}{2} = \frac{g(t_1 + t_2)^2}{8}\). Thus, the correct answer is (b).

Key Concepts

Vertical Motion Equation

When a body is projected vertically upward, it follows a unique path described by a set of mathematical principles known as the vertical motion equation. The key formula here is \[ h = ut - \frac{1}{2}gt^2 \]where:

• \( h \) is the height at a given time.
• \( u \) is the initial velocity of the projectile (i.e., how fast the object is going when first launched).
• \( g \) represents the acceleration due to gravity, which is typically around 9.8 m/s² downwards.
• \( t \) is the time elapsed since the object was projected.

This equation reveals how the initial velocity and gravitational pull affect the displacement and time of travel.
Understanding this formula is essential to solving problems related to projectile motion, as it helps us predict where and when an object will reach a particular height during its trajectory.
The projectile uses this path until it reaches maximum height and starts descending under gravity, maintaining symmetrical properties.

Maximum Height

The concept of maximum height in projectile motion refers to the highest vertical position reached by an object during its flight. To find this, we need to understand when the object's velocity along the vertical direction becomes zero. At the maximum height, all the vertical kinetic energy is converted into potential energy.
For the solution given, the critical moment happens when time equals half of the total time to cross a certain height, specifically when:\[ t = \frac{t_1 + t_2}{2} \]Here, we utilize the fact that at maximum height, the vertical velocity is zero:

• The initial velocity equation is rearranged to \( u = g\left(\frac{t_1 + t_2}{2}\right) \).
• The formula for maximum height can be derived from energy conservation principles: \[\text{Max Height} = \frac{u^2}{2g} = \frac{\left(g\left(\frac{t_1 + t_2}{2}\right)\right)^2}{2g}\]
• After simplification, this gives us the expression \[\frac{g(t_1 + t_2)^2}{8}\]

Reaching the maximum height is a turning point in the motion of the projectile since it is the point where the object stops rising and begins to fall back toward its starting height.

Symmetrical Projectile Motion

Symmetrical projectile motion is a fascinating aspect of physics that addresses how an object behaves under a uniform gravitational field without any air resistance. This symmetry implies that the time to ascend to the highest point is equal to the time to descend back to the original height point, given constant acceleration due to gravity.

• In the exercise above, the body reaches the same height \( h \) at two different times, \( t_1 \) and \( t_2 \).
• The overall time, \( t_1 + t_2 \), symbolizes the entire journey up to the max height and back to the point.
• The midpoint of this time span, or \( \frac{t_1 + t_2}{2} \), precisely marks the moment the body hits its peak height.

This symmetrical nature provides a straightforward way to anticipate the behavior of projectiles when monitoring their trajectory in a frictionless environment. It also facilitates the analysis by reducing the complexity of calculations involved, making it easier to solve projectile-related problems.