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20. A body is allowed to fall from a height of 98 m beforehitting the ground the distance travelled by it in the lastsecond of motion (9 = 9.8 m/s') is(a) 38.91 m (b) 40 m (c) 50 m (d) 29.91 m​

Answer»

answer : option (a) 38.91m

explanation : time taken by body to reach the ground , t = √{2h/g}

here, h = 98 m , g = 9.8 m/s²

so, t = √{2 × 98/9.8} = √(20) sec

so, we have to find DISTANCE travelled by PARTICLE during (√20 -1)s to √20 s.

velocity of body at t = (√20 - 1)s

v = u + at

here, u = 0, a = -g, t = √20 - 1

so, v = -g(√20 - 1) m/s

now, displacement travelled in LAST second, s = ut + 1/2 at²

here, u = v = -g(√20 - 1) , t = 1S, a = -g

so, s = -g(√20 - 1) - g/2

= -g[√20 - 1 + 0.5 ]

= -9.8 [4.47 - 0.5 ]

= -38.91 m [ here negative sign indicates that displacement of body is in downward direction]

hence, distance travelled by it in the last second = 38.91m



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