Asked by iqra
A stone is dropped from peak of the hill. It covers a distance of 30 metres in the last second of its motion. Find the height of the peak?
Answers
Answered by
Henry
h1 = V*t + 0.5g*t^2 = 30 m.
V*1 + 4.9*1^2 = 30
V = 30 - 4.9 = 25.1 m/s
h = (V^2-Vo^2)/2g + h1
h = (25.1^2-0)/19.6 + 30 = 62.14 m. = Ht. of the peak.
V*1 + 4.9*1^2 = 30
V = 30 - 4.9 = 25.1 m/s
h = (V^2-Vo^2)/2g + h1
h = (25.1^2-0)/19.6 + 30 = 62.14 m. = Ht. of the peak.
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