Question

A stone dropped from the peak of the hill it covers a distance of 30 metres in the last second of its motion find the height of the peak peak of the hill

Answer 1

Answer:

S1 = 1/2 g t1^2       distance fallen in time t1

S2 = 1/2 g t2^2      distance fallen in time t2

S2 - S1 = 30 = 1/2 g (t2^2 - t1^2)

60/g = t2^2 - t1^2 = (t1 + t2) * (t2 - t1) = t1 + t2     since t2 - t1 = 1

t1 + t2 = 6.12

t2 - 1 + t1 = 6.12         since t2 - t1 = 1

2 t1 = 7.12  Then

t2 = 3.56       and     t1 = 2.56

Check:

1/2 * 9.8 * 2.56^2 = 32.1

1/2 * 9.8 * 3.56^2 = 62.1

and Height = 62.1 m