A ball is dropped from the top of the tower of height h.it covers a distance h/2 in the last second of its motion.how long does the ball remain in air?
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For total journey
H = 0.5gT^2 ……[1]
For 1st half journey
H/2 = 0.5g (T - 1)^2 ……[2]
Divide equation [1] with [2]
H / (H/2) = [T / (T - 1)]^2
2 = [T / (T - 1)]^2
√2 = T / (T - 1)
T√2 - √2 = T
T (√2 - 1) = √2
T = √2 / (√2 - 1)
T = 1.414 / 0.414
T = 3.4 s
Ball will remain in air for 3.4 seconds
H = 0.5gT^2 ……[1]
For 1st half journey
H/2 = 0.5g (T - 1)^2 ……[2]
Divide equation [1] with [2]
H / (H/2) = [T / (T - 1)]^2
2 = [T / (T - 1)]^2
√2 = T / (T - 1)
T√2 - √2 = T
T (√2 - 1) = √2
T = √2 / (√2 - 1)
T = 1.414 / 0.414
T = 3.4 s
Ball will remain in air for 3.4 seconds
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