A ball is dropped from a certain height such that it covers a distance...
Let 'n' denote time in this case.w.k.t. S(n) = u + g/2(2n-1)44.1=0+9.8/2 (2n-1)44.1÷4.9=2n-19+1=2nn=5total time of fall=n=5.
also w.k.t.S=ut+g/2 (t^2)S=0+9.8/2 (5^2)S=4.9 (25)S=122.5mTotal height =122.5m
A ball is dropped from a certain height such that it covers a distance...
Calculating Total Height and Total Time of Fall
Given:
- Distance covered in the last second = 44.1m
Calculating Total Height:
- The distance covered in the last second of motion is the sum of the distance traveled in the last second before coming to rest.
- Using the formula for distance covered in the last second: \( s = u + (v - u) \times t \), where:
- \( s = 44.1m \) (distance covered in the last second)
- \( u = 0 \) (initial velocity)
- \( v = 0 \) (final velocity)
- \( t = 1s \) (time taken in the last second)
- Substituting the values, we get: \( 44.1 = 0 + (0 - 0) \times 1 \)
- Therefore, the total height of the drop = 44.1m
Calculating Total Time of Fall:
- To calculate the total time of fall, we need to consider the time taken for the ball to cover the total height.
- Using the equation of motion: \( s = ut + \frac{1}{2}at^2 \), where:
- \( s = 44.1m \) (total height)
- \( u = 0 \) (initial velocity)
- \( a = 9.81ms^{-2} \) (acceleration due to gravity)
- \( t \) is the total time of fall
- Substituting the values, we get: \( 44.1 = 0 + \frac{1}{2} \times 9.81 \times t^2 \)
- Solving for \( t \), we find: \( t = \sqrt{\frac{2 \times 44.1}{9.81}} \approx 3s \)
- Therefore, the total time of fall = 3 seconds
In conclusion, the total height of the drop is 44.1m and the total time of fall is 3 seconds.
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