A particle when thrown, moves such that it passes from same height at 2 and 10s, the height is
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Answered by
82
If t1 and t2 are the time when body is at the same height
Then
h = ½ * g* t1 * t2
h = ½* g * 2 * 10
h = 10 g
Answered by
117
answer : 100m
a/c to question,
it passes same height at 2s and 10s. Let us consider that a particle is thrown upwards with speed u m/s. so, trajectory of particle follows , s = ut + 1/2(-g)t² ,
particle reaches same height s, at t = 2s and t = 10s.
putting t = 2s
s = u(2) + 1/2(-g)(2)² = 2u - 2g ....(1)
putting t = 10s
s = u(10) + 1/2(-g)(10)² = 10u - 50g.....(2)
from equations (1) and (2),
s - s = 10u - 50g - (2u - 2g)
or, 0 = 8u -(50g - 2g)
or, 0 = 8(u - 6g)
or, u = 6g ......(3)
now, height , s = 2u - 2g = 2(6g) - 2g
= 10g = 100m [ if g = 10m/s² ]
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