Question

(30) A freely falling particle covers a building of 45m height in one second. Find the height of the

point from where the particle was released. [g = 10ms–2]

(A) 120m

(B) 125m

(C) 25m (D) 80m​

Answer 1

Explanation:

Let particle is thrown from A

BC = 45 m

at B let velocity be v1

∴ velocity while covering distance of building

V = V1 + at

V = V1 + (10 × 1)

V = V1 + 10

also V2 = V12 + 2ad

(V1 + 10)2 = V12 + (2 × 10 × 45)

V12 + 20V1 + 100 = V12 + 900

20V1 = 800

V1 = 40 m/s

V = 50 m/s

considering portion AB,

V1 = V0 + at

40 = 0 + (10)t

t = 4 sec

Hence total time by particle = 4 sec to cover distance AB + 1 sec to cover distance BC

= 5 sec

d = Vot + (1/2) at2

∴ d = (0) + (1/2) (10) (5)2

d = 125 m

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