A stone is dropped from the top of a cliff and is found to travel 44.1 m in the last second before it reaches the ground. Find the height of the cliff.
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uvik118 Ace
initial velocity(u ) = 0using formula for displacement in t th second
⇒ 44.1 = 1/2 x 9.8 (2t-1)⇒(44.1 x 2)9.8 + 1 = 2t⇒10/2 = t⇒t = 5 sec
so using equation
⇒s = 0 + 1/2 x 9.8 x (5)²⇒s = 4.9 x 25⇒s = 122.5 mso total height = 122.5
initial velocity(u ) = 0using formula for displacement in t th second
⇒ 44.1 = 1/2 x 9.8 (2t-1)⇒(44.1 x 2)9.8 + 1 = 2t⇒10/2 = t⇒t = 5 sec
so using equation
⇒s = 0 + 1/2 x 9.8 x (5)²⇒s = 4.9 x 25⇒s = 122.5 mso total height = 122.5
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Answer:
122.5 m
Explanation:
initial velocity(u ) = 0
using formula for displacement in t th second
⇒ 44.1 = 1/2 x 9.8 (2t-1)
⇒(44.1 x 2)9.8 + 1 = 2t
⇒10/2 = t
⇒t = 5 sec
so using equation
⇒s = 0 + 1/2 x 9.8 x (5)²
⇒s = 4.9 x 25
⇒s = 122.5 m
so total height = 122.5
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