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
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.
To make sure you are not studying endlessly, EduRev has designed Class 9 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 9.