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 :
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:
We need to find the horizontal distance \(R\) the stone travels before hitting the ground.
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.
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:
\(R = 12 \text{ m/s} \times 2 \text{ s}\)
\(R = 24 \text{ m}\)
The horizontal distance \(R\) from the building where the stone hits the ground is 24 m.
Comparing this result with the given options:
Our calculated value \(R = 24\) m matches Option 3.
| 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) | \(v_{xi} = v_{throw}\) | \(v_{yi} = 0\) |
Understanding projectile motion is key to solving many physics problems. Here are some extra points:
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:
The ration of the number of girls and boys in a school is 8 : 7. If the percentage increase in the number of girls and boys is 10% and 20% respectively, what will be the new ratio?
The cheapest means of transport is:
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Which newspaper edited by Bal Gangadhar Tilak, was one of the strongest critics of the British rule?
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