Question

Stones are thrown horizontally with the same velocity from two buildings. One stone lands twice as far away from its building as the other stone. Determine the ratio of the heights of the two buildings.

Short answer

Answer: The ratio of the heights of the two buildings is 1:2.

Step-by-step solution

Identify given information and the variables to find

We are given: - Two stones thrown horizontally with the same velocity (v) - One stone lands twice as far away from its building as the other (2x and x distance respectively) - We need to find the ratio of the heights of the buildings (h1/h2)

Calculate time of flight for both stones

To calculate the time of flight(t), we will use horizontal projectile motion formula: Horizontal distance = horizontal velocity × time x = vt t = x/v For stone 1: t1 = x/v For stone 2: t2 = 2x/v t2 = 2t1

Calculate the height of buildings using vertical displacement formula

We will use vertical displacement formula to find the heights (h1 and h2) of the buildings for stone 1 and stone 2: Vertical displacement = 1/2 × g × t^2 Where, g is the gravitational acceleration (9.81 m/s^2) For stone 1: h1 = 1/2 × g × t1^2 For stone 2: h2 = 1/2 × g × t2^2

Calculate the ratio of the heights of the buildings

Since we know t2 = 2t1, we can replace t2 in the formula for h2: h2 = 1/2 × g × (2t1)^2 h2 = 1/2 × g × 4t1^2 h2 = 2 × (1/2 × g × t1^2) Now, we can write h1/h2 as follows: h1/h2 = (1/2 × g × t1^2) / (2 × (1/2 × g × t1^2)) h1/h2 = (1/2 × g × t1^2) / (g × t1^2) h1/h2 = 1/2 Thus, the ratio of the heights of the two buildings is 1:2.

Key concepts

Horizontal Projectile Motion

Horizontal projectile motion is a form of motion where an object moves in a horizontal path while simultaneously being acted upon by gravity. In a scenario where objects are thrown with the same horizontal velocity from different heights, their horizontal velocity remains constant throughout their flight, since there are no horizontal forces acting on the object after it is released (assuming air resistance is negligible).

The horizontal distance traveled by the object is simply the product of its horizontal velocity and the time it spends in the air – this is known as the 'time of flight'. This concept is crucial for solving problems involving objects launched horizontally from heights, as it allows us to determine how far the objects will land from their initial points.

In the given exercise, two stones are thrown with the same horizontal velocity from buildings of different heights. To find the ratio of the heights, we need to understand that the travel distance of each stone is directly linked to the time of flight for that stone, given the horizontal velocity is constant for both.

Time of Flight

Time of flight in the context of projectile motion refers to the duration an object remains in the air before hitting the ground. It's a critical factor for determining an object's trajectory and landing point. To understand the time of flight, one must bear in mind that it is independent of horizontal velocity when projecting horizontally.

For horizontal projectile motion, time of flight is determined by the vertical displacement of the object. Since gravity is the only vertical force acting on the object, the time it takes to fall from its original height to ground level is purely dependent on this vertical distance and the acceleration due to gravity, which on Earth is approximately 9.81 m/s².

As illustrated in the solution, by knowing the horizontal distance each stone travels (x for the first stone and 2x for the second stone), and the constant horizontal velocity, we can deduce the time of flight for both stones. Remarkably, the second stone's time of flight is double that of the first, indicating twice the duration in air, which in turn plays a vital role in determining the vertical height from which each stone was released.

Vertical Displacement

Vertical displacement in projectile motion is the measure of how far an object has moved in the vertical direction during its flight. When an object is thrown horizontally, the only force acting on it in the vertical direction is gravity. This means the vertical displacement is a result of the object accelerating downwards due to gravity, starting from an initial vertical velocity of zero.

The formula for vertical displacement, \( s = \frac{1}{2} g t^2 \), allows us to calculate the height from which an object falls if we know the time it spent in the air (time of flight). Here, 'g' stands for the acceleration due to gravity and 't' for time. This quadratic relationship implies that if the time of flight doubles, as in the case of the second stone, the vertical displacement (or height in this context) will be quadrupled, assuming a constant acceleration due to gravity.

Understanding this concept is essential in our exercise, as we see that although the second stone was in the air for twice as long as the first, the resulting vertical displacement is not merely double, but four times greater, which leads to the ratio of the heights of the two buildings being 1:2.