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A coin is dropped from a tower. It moves to through a distance 24.5m in the last second before hitting the ground .find the height of the tower?
Most Upvoted Answer
A coin is dropped from a tower. It moves to through a distance 24.5m i...
Distances travelled in last second,
S = u + a(n - 1/2),
24.5 = 0 + 9.8(n - 1/2),
n = 24.5/9.8 + 0.5,
n = 3rd second,

Total time of flight is 3 seconds,

Height of the tower
H = 0.5gt^2,
= 0.5 × 10 × 3^2,
= 45 m
Community Answer
A coin is dropped from a tower. It moves to through a distance 24.5m i...



Given Data:

- Distance covered in the last second before hitting the ground = 24.5m

Formula:

- The distance covered by an object in free fall during the last second before hitting the ground can be given by the formula:
\[d = gt^2\]
where:
d = distance covered in the last second (24.5m)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken to cover the distance

Calculations:

- Given that the distance covered in the last second is 24.5m, we can rearrange the formula to solve for time:
\[24.5 = 9.8t^2\]
\[t^2 = \frac{24.5}{9.8}\]
\[t^2 = 2.5\]
\[t = \sqrt{2.5}\]
\[t = 1.58s\]

- Now, we can find the height of the tower using the formula for distance covered in free fall:
\[h = \frac{1}{2}gt^2\]
\[h = \frac{1}{2} \times 9.8 \times (1.58)^2\]
\[h \approx 12.4m\]

Conclusion:

- The height of the tower is approximately 12.4m.
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A coin is dropped from a tower. It moves to through a distance 24.5m in the last second before hitting the ground .find the height of the tower?
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