A stone is released from the top of a tower 125 m high. Then find:(1)

1.	A stone is released from the top of a tower 125 m high. Then find:(1) the time with which the stone strikes the ground.(2) the final velocity of the stone. (Take g = 10 m/s2)
Answer» We are given that: Height of tower (h) = 125 m Initial velocity (u) = 0 g = 10 m/s² (1) From 2nd equation of motion For downward motion: h = ut + \(\frac{1}{2}\)gt² ⇒ 125 = 0 x t + \(\frac{1}{2}\) x 10 x t² = 125 = 5t² ⇒ t² = 25 ⇒ t = 5sec² (2) From 1st equation of motion, v = u + gt ⇒ u = 0 + 10 x 5 ⇒ v = 50 m/s

A stone is released from the top of a tower 125 m high. Then find:(1) the time with which the stone strikes the ground.(2) the final velocity of the stone. (Take g = 10 m/s2)

Answer»

We are given that: Height of tower (h) = 125 m

Initial velocity (u) = 0

g = 10 m/s²

(1) From 2nd equation of motion

For downward motion: h = ut + \(\frac{1}{2}\)gt²

⇒ 125 = 0 x t + \(\frac{1}{2}\) x 10 x t² = 125 = 5t²

⇒ t² = 25

⇒ t = 5sec²

(2) From 1st equation of motion,

v = u + gt

⇒ u = 0 + 10 x 5

⇒ v = 50 m/s

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A stone is released from the top of a tower 125 m high. Then find:(1) the time with which the stone strikes the ground.(2) the final velocity of the stone. (Take g = 10 m/s2)

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