A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of `65^(@)` with the horizontal. From what height above the ground was the marble thrown?

Options

A. 19 m
B. 38 m
C. 53 m
D. 47 m

Answer

To solve the problem, we need to determine the height from which the marble was thrown horizontally. Let's break down the solution step by step.

### Step 1: Understand the motion of the marble
The marble is thrown horizontally with an initial speed of 15 m/s. It experiences two types of motion:
- Horizontal motion with constant velocity (15 m/s).
- Vertical motion under the influence of gravity.

### Step 2: Analyze the final velocity components
When the marble strikes the ground, it has a velocity that makes an angle of 65 degrees with the horizontal. We can denote:
- \( V_x \) = horizontal component of velocity = 15 m/s (constant)
- \( V_y \) = vertical component of velocity (to be determined)

Using the angle given, we can relate the vertical and horizontal components of the velocity:
\[ \tan(65^\circ) = \frac{V_y}{V_x} \]
Substituting \( V_x = 15 \, \text{m/s} \):
\[ \tan(65^\circ) = \frac{V_y}{15} \]
Thus, we can find \( V_y \):
\[ V_y = 15 \cdot \tan(65^\circ) \]

### Step 3: Calculate \( V_y \)
Using a calculator:
\[ \tan(65^\circ) \approx 2.1445 \]
So,
\[ V_y = 15 \cdot 2.1445 \approx 32.17 \, \text{m/s} \]

### Step 4: Use kinematic equations to find height
We can use the kinematic equation for vertical motion to find the height \( h \):
\[ V_y^2 = V_{y0}^2 + 2gh \]
Since the marble is thrown horizontally, the initial vertical velocity \( V_{y0} = 0 \):
\[ V_y^2 = 0 + 2gh \]
Rearranging gives:
\[ h = \frac{V_y^2}{2g} \]
Substituting \( V_y \approx 32.17 \, \text{m/s} \) and \( g \approx 9.8 \, \text{m/s}^2 \):
\[ h = \frac{(32.17)^2}{2 \cdot 9.8} \]
Calculating \( (32.17)^2 \):
\[ (32.17)^2 \approx 1032.43 \]
Now substituting:
\[ h = \frac{1032.43}{19.6} \approx 52.66 \, \text{m} \]

### Step 5: Final answer
Rounding to the nearest meter, the height from which the marble was thrown is approximately:
\[ h \approx 53 \, \text{m} \]

### Summary
The marble was thrown from a height of approximately **53 meters**.

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