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 :
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.
Important Questions from Miscellaneous
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be:
