Question:
A ball is thrown upward from the top of a building at an angle of \(30^\circ\) to the horizontal with an initial speed of 15 m/s. If the ball hits the ground after 3 s, find the height of the building. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
A ball is thrown upward from the top of a building at an angle of \(30^\circ\) to the horizontal with an initial speed of 15 m/s. If the ball hits the ground after 3 s, find the height of the building. (Acceleration due to gravity \(g = 10 \, \text{m/s}^2\))
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For projectile motion from a height, always resolve velocity into vertical and horizontal components, then use \(y = u_y t - \frac{1}{2} g t^2\) to find vertical displacement and building height.
Updated On: Jul 18, 2026
- 30 m
- 12.5 m
- 25.5 m
- 22.5 m
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The Correct Option is D
Solution and Explanation
Step 1: Resolve the initial velocity.
Initial velocity \(u = 15 \, \text{m/s}\) is at an angle \(\theta = 30^\circ\) to the horizontal. The vertical component is:
\[ u_y = u \sin \theta = 15 \times \frac{1}{2} = 7.5 \, \text{m/s} \]
The horizontal component is \(u_x = u \cos \theta = 15 \times \frac{\sqrt{3}}{2}\), but horizontal motion is not needed for height.
Step 2: Use vertical motion equation.
For vertical displacement \(y\) from the top of the building:
\[ y = u_y t + \frac{1}{2}(-g)t^2 \]
Here, \(y\) is negative since ball hits the ground below initial position, \(t = 3 \, \text{s}\), \(g = 10 \, \text{m/s}^2\):
\[ y = 7.5 \times 3 - \frac{1}{2} \cdot 10 \cdot 3^2 \]
Step 3: Calculate vertical displacement.
\[ y = 22.5 - 45 = -22.5 \, \text{m} \]
Negative sign indicates the ball lands 22.5 m below the launch point.
Step 4: Interpret result.
Since the ball hits the ground, the magnitude of displacement corresponds to the height of the building:
\[ h = 22.5 \, \text{m} \]
Step 5: Verify using motion concepts.
The time of flight \(t = 3\) s is consistent with vertical motion under gravity. The vertical initial velocity component \(u_y = 7.5 \, \text{m/s}\) and deceleration \(g = 10 \, \text{m/s}^2\) yield the correct height.
Step 6: Final conclusion.
Thus, the height of the building is:
\[ \boxed{22.5 \, \text{m}} \]
Initial velocity \(u = 15 \, \text{m/s}\) is at an angle \(\theta = 30^\circ\) to the horizontal. The vertical component is:
\[ u_y = u \sin \theta = 15 \times \frac{1}{2} = 7.5 \, \text{m/s} \]
The horizontal component is \(u_x = u \cos \theta = 15 \times \frac{\sqrt{3}}{2}\), but horizontal motion is not needed for height.
Step 2: Use vertical motion equation.
For vertical displacement \(y\) from the top of the building:
\[ y = u_y t + \frac{1}{2}(-g)t^2 \]
Here, \(y\) is negative since ball hits the ground below initial position, \(t = 3 \, \text{s}\), \(g = 10 \, \text{m/s}^2\):
\[ y = 7.5 \times 3 - \frac{1}{2} \cdot 10 \cdot 3^2 \]
Step 3: Calculate vertical displacement.
\[ y = 22.5 - 45 = -22.5 \, \text{m} \]
Negative sign indicates the ball lands 22.5 m below the launch point.
Step 4: Interpret result.
Since the ball hits the ground, the magnitude of displacement corresponds to the height of the building:
\[ h = 22.5 \, \text{m} \]
Step 5: Verify using motion concepts.
The time of flight \(t = 3\) s is consistent with vertical motion under gravity. The vertical initial velocity component \(u_y = 7.5 \, \text{m/s}\) and deceleration \(g = 10 \, \text{m/s}^2\) yield the correct height.
Step 6: Final conclusion.
Thus, the height of the building is:
\[ \boxed{22.5 \, \text{m}} \]
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