A ball is dropped from the roof of a tower of height h. The total distance covered it in the last seconds of its motion is equal to the distance covered by it in first three seconds. The value of h in meters is $(g = 10 m / s^{2})$

125

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200

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100

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Solution

The correct option is A 125
A ball is dropped from the roof of a tower height $= h$
Total distance covered it in the last seconds of its motion
Solution
The distance covered in first $3$ second
$s = \frac{1}{2} a t^{2} = \frac{1}{2} \times 10 \times 3^{2} = 45$
If the ball takes in second to fall to ground. The distance covered in $n t h$ second
$s_{n} = u + \frac{g}{2} (2 n - 1) = 0 + \frac{10}{2} (2 \times n - 1) = 10 n - 5$
Therefore, $45 = 10 n - 5 10 n = 50 n = 5$
Therefore $h = \frac{1}{2} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 25 = 125 m$

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A ball is dropped from the roof of a tower of height h. The total distance covered it in the last seconds of its motion is equal to the distance covered by it in first three seconds. The value of h in meters is (g=10m/s2)

A ball is dropped from the roof of a tower of height h. The total distance covered it in the last seconds of its motion is equal to the distance covered by it in first three seconds. The value of h in meters is $(g = 10 m / s^{2})$