A ball thrown upward from the top of tower with speed V reaches the ground in T1 second. If this ball is thrown downward from the top of the same tower with speed V it reaches the ground in T2 second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from top of the tower? (a) T1+T2/2 (b) T2-T1/2 (c) √T1T2 (d) T1+T2 Solve this Question !!
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Answer:
as ɨt ɨs ɢɨʋɛռ tɦat tɦɛ sքɛɛɖ ɨs ʋ. sօ, ɮʏ ʊsɨռɢ sɛċօռɖ ɛզʊatɨօռ օʄ ҡɨռɛʍatɨċs
s = ʊt + 1/2 at²
ɮʊt ɦɛʀɛ,
a = ɢ
ʊ = ʋ
s = s
s = ʋt₂ + 1/2 × ɢ × (t2)²
sɨռċɛ, tɦɛ ɮaʟʟ ɨs tɦʀօառ ʊքաaʀɖ. sօ, tɨʍɛ taҡɛռ աɦɛռ tɦɛʏ աɨʟʟ ċօʍɛ at saʍɛ քօɨռt = t1 - t2
tɨʍɛ taҡɛռ ɮʏ t1 ʄօʀ ʀɛaċɦɨռɢ tօք
t1/2
tɨʍɛ taҡɛռ ɮʏ t2 ʄօʀ ʀɛaċɦɨռɢ tօք
t2/2
tɨʍɛ taҡɛռ tօ ċօʍɛ at tօք ɨռċʟʊɖɨռɢ ɮօtɦ ʊքաaʀɖ aռɖ ɖօառաaʀɖ\
t1/2 - t2/2
t1 - t2/2
as աɦɛռ ɨt ċօʍɛ at tօք. tɦɛռ, ɨts ʋɛʟօċɨtʏ աɨʟʟ ɮɛċօʍɛ 0
s = ʋt₂ + 1/2 × ɢ × (t2)²
0 = ʋ - ɢ × (t1/2- t2/2)
0 = ʋ - ɢ × (t1 - t2/2)
0 + ʋ = ɢ(t1 - t2/2)
ʋ = ɢ(t1 - t2/2)
ռօա
s = t2ɢ(t1 - t2/2) + 1/2 × ɢ × t2²
s = t2ɢ(t1 - t2/2) + ɢ/2 × t2²
taҡɨռɢ ɢ aռɖ t2 as ċօʍʍօռ
s = ɢ × t1 × t2/2 (1)
s = ɢ × t²/2
s = 1/2 × ɢ × t²
s = 1/2 ɢt² (2)
օռ ċօʍքaʀɨռɢ ɮօtɦ
ɢ × t1 × t2/2 = 1/2 × ɢt²
ɢ × t²/2 = 1/2 × ɢt²
t²/2 = 1/2 × t²
t² = t²
t = √(t1t2)
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