a body is released from height h above the ground exactly at the mid if g vanishes suddenly total time of fall is
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1) For traveling a height of h/2 from rest the time taken is:
h/2 = 0 * t1 + 1/2 g t1²
=> t1 = √(h / g)
Speed of the body at this height => v = 0 + g * √(h/g) = √(g h)
2) For traveling the remaining height of h/2 at the above uniform speed:
h/2 = √(g h) * t2
=> t2 = 1/2 * √(h / g)
Total time taken to reach the ground: 3/2 * √(h / g)
h/2 = 0 * t1 + 1/2 g t1²
=> t1 = √(h / g)
Speed of the body at this height => v = 0 + g * √(h/g) = √(g h)
2) For traveling the remaining height of h/2 at the above uniform speed:
h/2 = √(g h) * t2
=> t2 = 1/2 * √(h / g)
Total time taken to reach the ground: 3/2 * √(h / g)
kvnmurty:
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Answer:
3/2 root h/g sec
Explanation:
1) For traveling a height of h/2 from rest the time taken is:
h/2 = 0 * t1 + 1/2 g t1²
=> t1 = √(h / g)
Speed of the body at this height => v = 0 + g * √(h/g) = √(g h)
2) For traveling the remaining height of h/2 at the above uniform speed:
h/2 = √(g h) * t2
=> t2 = 1/2 * √(h / g)
Total time taken to reach the ground: 3/2 * √(h / g)
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