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A cubical box dimension L = 5//4 m start...

A cubical box dimension `L = 5//4` m starts moving with an acceleration `veca = 0.5 ms^(-2) hati` form the state of rest. At the same time, a stone is thrown form the origin with velocity `vecV = v_1hati + v_2hatj - v_3hatk` with respect to earth. Acceleration due to gravity `vecg = 10ms^(-2)(-hatj)`. The stone just touches the roof of box and finally falls at the diagonally opposite point. then:

A

`v_1 = 3/2`

B

`v_2 = 5`

C

`v_3 = 5/4`

D

`v_3 = 5/2`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

a,b,c. `H = (u_(2)^(y))/(2g) rArr u_y = V_2 = 5ms^(-1)`
`T = (2usintheta)/g = (2u_y)/g = 1s`
`L = u_xt - 1/2 a_xt^2`
`5/4 = u_x xx 1 - 1/2 xx (0.5) xx (1)^2 rArr u_x = V_1 = 3//2 ms^(-1)`
Along `z` axis `5/4 = u_z xx 1` rArr `v_z = 5/4 ms^(-1)` .
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