A body is projected with a velocity of `20 ms^(-1)` in a direction making an angle of `60^(@)` with the horizontal. Determine its (i) position after 0.5 s and (ii) the velocity after `0.5s`.

Options

A. `15m` , `7.43 m` , `10 ms^(-1)`
, `12.42 ms^(-1)`
B. `5m` , `6.43 m` , `10 ms^(-1)`
, `12.42 ms^(-1)`
C. `5m` , `7.43 m` , `10 ms^(-1)`
, `12.42 ms^(-1)`
D. `15m` , `7.43 m` , `10 ms^(-1)`
, `1.42 ms^(-1)`

Answer

C

Explanation

Given, `u = 20 ms^(-1), theta = 60^(@), t = 0.5s`
(i) Since, horizontal distance,
`x = (u cos theta) t = (20 cos 60) xx 0.5 = 5m`
Similarly, vertical distance,
`y = (u sin theta) t -(1)/(2) g t^(2)`
`= (20 sin 60^(@)) xx 0.5 -(1)/(2) xx 9.8 xx (0.5)^(2) rArr y = 7.43 m`
(ii) Velocity along horizontal direction,
`v_(x) = u cos theta = 20 cos 60^(@) = 10 ms^(-1)`
Velocity along vertical direction,
`v_(y) = u sin theta - g t = 20 sin 60^(@) - 9.8 xx 0.5 = 12.42 ms^(-1)`