A ball is projected with a velocity of `20 m//s` vertically. Find the distance travelled in first three second. (use `g = 10 m//sec^(2)` )

Options

A. `20 m`
B. `25 m`
C. `30 m`
D. `50 m`

Answer

B

Explanation

Problem here is to find the distance. We can calculate that the direction of ball is changes at `t = 2s`. (From `v = u + at`, since `v = 0` at highest point therefore `0 = 20 - 10 t rArr t = 2s` )
Distance traveled in first two second (b) ( `:.` Distance = Displacement, because o f velocity does not change direction in one dimension)
`:. S_(f) = 20 xx 2 - (1)/(2) 10 xx4`
`40 - 20 = 20 m` (upward)
Distance traveled in next one seconds
`S_(2) = (1)/(2) 10 xx 1 = 5m` downward
So, total distance travelled by the ball in first three seconds (br) `20 + 5 = 25 m`