In the question number 52, the speed with which the stone hits the ground is

Options

A. 15 m `s^(-1)`
B. 90 m `s^(-1)`
C. 99 m `s^(-1)`
D. 49 m `s^(-1)`

Explanation

To find the speed with which the stone hits the ground, we need to consider both the horizontal and vertical components of the stone's motion.

### Step-by-Step Solution:

**Step 1: Analyze the horizontal motion.**
- The initial horizontal speed \( u_x \) is given as \( 15 \, \text{m/s} \).
- The horizontal acceleration \( a_x \) is \( 0 \, \text{m/s}^2 \) (no horizontal forces acting).
- The final horizontal speed \( v_x \) can be calculated using the formula:
\[ v_x = u_x + a_x t \]
Substituting the values:
\[ v_x = 15 + 0 \cdot 10 = 15 \, \text{m/s} \]

**Step 2: Analyze the vertical motion.**
- The initial vertical speed \( u_y \) is \( 0 \, \text{m/s} \) (the stone is dropped).
- The vertical acceleration \( a_y \) is the acceleration due to gravity \( g \), which is approximately \( 9.8 \, \text{m/s}^2 \).
- The final vertical speed \( v_y \) can be calculated using the formula:
\[ v_y = u_y + a_y t \]
Substituting the values:
\[ v_y = 0 + 9.8 \cdot 10 = 98 \, \text{m/s} \]

**Step 3: Calculate the resultant speed when the stone hits the ground.**
- The resultant speed \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(15)^2 + (98)^2} = \sqrt{225 + 9604} = \sqrt{9829} \]
Calculating this gives approximately:
\[ v \approx 99 \, \text{m/s} \]

Thus, the speed with which the stone hits the ground is approximately \( 99 \, \text{m/s} \).