A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance will give :

Question

Prepp Team · Accepted Answer

Calculating the Range of a Horizontally Thrown Stone from a Building

This problem involves analyzing the motion of a stone thrown horizontally from a height, which is a classic example of projectile motion. In projectile motion, we typically analyze the horizontal and vertical components of motion independently, assuming air resistance is negligible.

We are given:

Initial horizontal speed, \(v_x = 12\) m/s
Height of the building, \(h = 20\) m
Acceleration due to gravity, \(g = 10\) m/s\(^2\)

We need to find the horizontal distance \(R\) the stone travels before hitting the ground.

Analyzing the Vertical Motion to Find Time of Flight

The stone is thrown horizontally, which means its initial vertical velocity is zero (\(v_{yi} = 0\)). The vertical motion is governed by gravity, causing a constant downward acceleration (\(a_y = g = 10\) m/s\(^2\)). The stone falls a vertical distance equal to the height of the building, \(h = 20\) m.

We can use the following kinematic equation for vertical displacement:

\(\Delta y = v_{yi}t + \frac{1}{2}a_yt^2\)

Let's take the downward direction as positive. So, initial vertical velocity \(v_{yi} = 0\), vertical displacement \(\Delta y = h = 20\) m, and acceleration \(a_y = g = 10\) m/s\(^2\). Let \(t\) be the time taken to hit the ground.

\(20 = (0)t + \frac{1}{2}(10)t^2\)

\(20 = 5t^2\)

\(t^2 = \frac{20}{5}\)

\(t^2 = 4\)

Taking the square root, we get \(t = \pm 2\). Since time must be positive, the time taken for the stone to hit the ground is \(t = 2\) seconds.

Analyzing the Horizontal Motion to Find the Range R

Since air resistance is neglected, there is no horizontal acceleration (\(a_x = 0\)). The horizontal velocity remains constant throughout the motion. The initial horizontal speed is \(v_x = 12\) m/s.

The horizontal distance covered (the range \(R\)) is given by the product of the constant horizontal velocity and the time of flight (\(t\)):

\(R = v_x \times t\)

Using the values we found:

\(R = 12 \text{ m/s} \times 2 \text{ s}\)

\(R = 24 \text{ m}\)

Conclusion

The horizontal distance \(R\) from the building where the stone hits the ground is 24 m.

Comparing this result with the given options:

Option 1: R = 12 m
Option 2: R = 18 m
Option 3: R = 24 m
Option 4: R = 30 m

Our calculated value \(R = 24\) m matches Option 3.

Projectile Motion Revision Table

Aspect	Horizontal Component	Vertical Component
Acceleration (neglecting air resistance)	\(a_x = 0\)	\(a_y = g\) (downward)
Velocity	Constant (\(v_x = v_{xi}\))	Changes due to gravity (\(v_y = v_{yi} + gt\))
Displacement	\(x = v_{xi}t\)	\(y = v_{yi}t + \frac{1}{2}gt^2\)
Initial Velocity (Horizontal Throw)	\(v_{xi} = v_{throw}\)	\(v_{yi} = 0\)

Additional Information on Projectile Motion Concepts

Understanding projectile motion is key to solving many physics problems. Here are some extra points:

Trajectory: The path followed by a projectile in the absence of air resistance is a parabola.
Independence of Motion: The most important concept is that the horizontal and vertical motions are independent of each other. Gravity only affects the vertical motion, not the horizontal velocity (assuming no air resistance).
Time of Flight: The time the projectile spends in the air is determined purely by the vertical motion and the height it falls or rises. In this problem, the time to hit the ground is determined by the height of the building and gravity.
Range: The horizontal range is the total horizontal distance covered. It depends on the horizontal velocity and the time of flight. A longer time of flight or a larger horizontal velocity results in a greater range.
Assumptions: Problems like this typically neglect air resistance and the rotation of the Earth. These assumptions simplify the analysis significantly, allowing us to treat 'g' as constant.