Problem 18 - Master Resource Book in JEE Main Physics

Question

A body of mass \(m\) thrown horizontally with velocity \(v\), from the top of tower of height \(h\) touches the level ground at distance of \(250 \mathrm{~m}\) from the foot of the tower. A body of mass \(2 \mathrm{~m}\) thrown horizontally with velocity \(\frac{v}{2}\), from the top of tower of height \(4 h\) will touch the level ground at a distance \(x\) from the foot of tower. The value of \(x\) is (a) \(250 \mathrm{~m}\) (b) \(500 \mathrm{~m}\) (c) \(125 \mathrm{~m}\) (d) \(250 \sqrt{2} \mathrm{~m}\)

Step-by-Step Solution

Answer

The value of \( x \) is 250 meters.

Step 1: Calculate Time of Flight for First Body

For a body projected horizontally, the time taken to reach the ground depends only on the height of the tower. The formula for time of flight is \( t = \sqrt{\frac{2h}{g}} \), where \( g \) is the acceleration due to gravity. So, the time to hit the ground for the first body is given by \( t_1 = \sqrt{\frac{2h}{g}} \).

Step 2: Find Horizontal Range for First Body

The horizontal distance covered, or range, is given by \( R = v \cdot t \). For the first body, \( v \cdot \sqrt{\frac{2h}{g}} = 250 \). Thus, the horizontal range is 250 meters.

Step 3: Calculate Time of Flight for Second Body

The second body is thrown from a height of \( 4h \). Using the flight time formula again, \( t_2 = \sqrt{\frac{8h}{g}} = 2\sqrt{\frac{2h}{g}} = 2t_1 \).

Step 4: Find Horizontal Range for Second Body

The second body is thrown with velocity \( \frac{v}{2} \). Thus, the horizontal range for this body is \( R = \frac{v}{2} \times 2t_1 = v \cdot t_1 \). Since \( v \cdot t_1 = 250 \), the range \( R = 250 \) meters.

Step 5: Conclusion

The horizontal distance \( x \) from the foot of the tower for the second body is equal to the first body's range, which is \( 250 \) meters.

Key Concepts

Horizontal Velocity

When discussing projectile motion, **horizontal velocity** is a critical component. It refers to the speed at which an object moves along the horizontal axis. For projectiles, this is generally assumed to be constant because, without air resistance, no horizontal forces act upon the object after it's thrown.

• Initial horizontal velocity, often denoted as \(v\), is the speed at which the projectile is launched.
• In our exercise, a body is thrown with a horizontal velocity of \(v\) from a certain height, and its motion is considered purely horizontal before it hits the ground.

This constancy is why the formula for the horizontal range, \(R = v \cdot t\), is straightforward. The range depends on both the horizontal velocity and the time of flight. If the initial velocity is doubled or halved, as in the exercise, the horizontal distance affects accordingly while keeping time and other factors the same.

Time of Flight

The **time of flight** is the total time a projectile stays in the air before touching the ground. With no vertical component of the initial velocity, the time it takes to fall depends solely on the height from which it was dropped and the acceleration due to gravity.

• Time of flight formula: \( t = \sqrt{\frac{2h}{g}} \)
• Directly related to the height: Higher heights result in longer flight times.
• In the provided solution, we compute the time of flight for two different heights \(h\) and \(4h\).

The exercise illustrates that the second body, thrown from a greater height, has a longer flight duration, specifically twice that of the first body. This increased time allows for more horizontal motion, which affects how far the second object travels.

Range of Projectile

The **range of a projectile** is the horizontal distance it covers during its motion. Understanding the range involves considering both the time of flight and the horizontal velocity.

• Range formula: \(R = v \cdot t\)
• All else being equal, range increases with a higher horizontal velocity or a longer time of flight.
• In our exercise, the first body's range was specified directly as 250 meters.

For the second body, the exercise had us determine if its range was the same, greater, or lesser under different initial conditions. Despite a lower initial velocity, the increased time due to greater height equaled out the range to be exactly 250 meters, showcasing the balancing act between time and velocity in determining range.