Question

If an object is thrown horizontally with an initial speed 10 m s-1 from the top of a building of height 100 m the horizontal distance covered by the particle is 45m

• A. true
• B. false

Answer 1

Answer: (A)

true

Answer 2

The problem describes a projectile motion scenario where an object is thrown horizontally from a certain height. We need to determine if the stated horizontal distance covered is correct.

1. Identify Given Parameters:

• Initial horizontal speed, $u_x = 10 \text{ m/s}$
• Height of the building, $h = 100 \text{ m}$
• Initial vertical speed, $u_y = 0 \text{ m/s}$ (since it's thrown horizontally)
• Acceleration due to gravity, $g$. We will use $g = 10 \text{ m/s}^2$ for simplicity.

2. Calculate the Time of Flight (t): The vertical motion of the object is governed by the equation: $h = u_y t + \frac{1}{2} g t^2$ Since $u_y = 0$: $h = \frac{1}{2} g t^2$ Substitute the given values: $100 = \frac{1}{2} \times 10 \times t^2$ $100 = 5 t^2$ $t^2 = \frac{100}{5}$ $t^2 = 20$ $t = \sqrt{20} \text{ s}$ $t = 2\sqrt{5} \text{ s}$

Numerically, $\sqrt{5} \approx 2.236$, so $t \approx 2 \times 2.236 = 4.472 \text{ s}$.

3. Calculate the Horizontal Distance (Range, R): The horizontal motion is uniform (constant velocity) as there is no horizontal acceleration (neglecting air resistance). The horizontal distance covered is given by: $R = u_x t$ Substitute the values of $u_x$ and $t$: $R = 10 \times 2\sqrt{5}$ $R = 20\sqrt{5} \text{ m}$

Numerically, using $t \approx 4.472 \text{ s}$: $R \approx 10 \times 4.472$ $R \approx 44.72 \text{ m}$

4. Compare with the Given Statement: The statement claims the horizontal distance covered by the particle is 45 m. Our calculated value is approximately 44.72 m. This is very close to 45 m. Therefore, the statement is considered true.