A body is projected horizontally from the top of a tower of height 180m with a velocity of `20 ms^(-2)`. If acceleration due to gravity is `10 ms^(-1)`, then match the following.

Options

A. `{:(A,B,C,D),(IV,II,III,I):}`
B. `{:(A,B,C,D),(I,II,III,IV):}`
C. `{:(A,B,C,D),(IV,II,I,III):}`
D. `{:(A,B,C,D),(II,IV,I,III):}`

Answer

C

Explanation

Given, initial horizontal component of velocity,
`u_(x) = 20 m//s`
Initial vertical component of velocity,
`u_(y) = 0`
Acceleration due to gravity,
`g = a_(y) = 10 m//s^(2)`
So, horizontal component of velocity after 1s,
`v_(x) = u_(x) + a_(x)t`
`=20 + 0 = 20 m//s`
and vertical component of velocity after 1s, `v_(y) = u_(y) + a_(y) t`
`= 0 + 10 xx 1`
=10 m/s
Horzontal displacement after 1s,
`s_(x) = u_(x) t + (1)/(2) a_(x) t^(2)`
`=20 xx 1 + 0 = 20m`
Vertical displacement after 1s, `s_(y) = u_(y) t + (1)/(2) a_(y) t^(2)`
`=0 + (1)/(2) xx 10 xx (1)^(2) = 5m`
Resultant velocity after 1s,
`v = sqrt(v_(x)^(2) + v_(y)^(2))`
`= sqrt((20)^(2) + (10)^(2)) = sqrt(400 + 100) = sqrt500`
= 22.36 = 22.4 m/s