A 0.5 kg ball falls past a window, whose top is 1.8 m in vertical extent from the ground. If the speed of the ball at the top of the window is 8 m/s, its speed at the ground will be : (Take g = 10 m/s²)
- AOption (a)
10 m/s
- BOption (b)
8 m/s
- COption (c)
6 m/s
- DOption (d)
5 m/s
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Correct answer
10 m/s
Why this answer is correct
Concept:
We can use the equations of motion to find the speed of the ball at the ground.
The ball is subject to gravity, and as it falls, its speed increases.
The final speed depends on the initial speed, the distance fallen, and the acceleration due to gravity.
We use the equation of motion: v<sub>2</sub> = u2 + 2gh
Where:
- v is the final velocity (speed at the ground),
- u = 8 m/s is the initial velocity (speed at the top of the window),
- h = 1.8 m is the distance fallen from the top of the window to the ground,
- g = 9.8 m/s2 is the acceleration due to gravity.
Calculation:
Here,
- Mass of the ball m = 0.5 kg (not needed for this calculation)
- Initial speed at the top of the window, v1 = 8 m/s
- Height of the top of the window from the ground, h = 1.8 m
- Acceleration due to gravity g = 9.8 m/s2
\( v^2 = (8)^2 + 2 \times 9.8 \times 1.8 \)
\( v^2 = 64 + 2 \times 9.8 \times 1.8 \)
v2 = 64 + 35.28
v2 = 99.28
Take the square root of both sides to find v,
\( v = \sqrt{99.28} \approx 9.96 , \text{m/s} \) (closer to 10m/s)
The correct answer is option (1).