A stone is thrown horizontally with velocity u. The velocity of the stone 0.5 s later is 3u/2. The value of u is

Options

A. 2.2 m/s
B. 3.3 m/s
C. 4.4 m/s
D. 5.5 m/s

Answer

C

Explanation

At t = 0.5 s, the vertical component of velocity is
`v_t = "gt" = g (0.5) = g/2 =4.9 m//s`
Hence at t = 0.5 s, we have
`sqrt(v_t^2+u^2)=(3u)/2 " or " (4.9)^2+u^2=(9u^2)/4`
or `5u^(2) = 4 xx (4.9)^(2) " or " u^2 = (0.8) (4.9)^2`
or `u = (4.9) sqrt(0.8) =4.4 ` m/s