A particle is fired horizontally with a velocity of `98 ms^(-1)` from the top of a hill 490m high, Find (i) the time taken to reach the ground (ii) the distance of the target from the hill and (iii) the velocity with which the projectile hits the ground `(g=9.8m//s^(2))`

Answer

(i) 10 sec, (ii) 980 m, (iii) `98sqrt(2) m//sec`

Explanation

`u_(y)=0, u_(x)=98 m//s`
on y-axis
`s= ut+1/2 at^(2)`
`490=0+ 1/2 xx 9.8 xx t^(2)`
`t=sqrt((490xx2)/9.8)=10 sec`
on x-axis for horizontal distance,
`R= u t- 98xx10=980 m`
`v_(x)=98 m//s`
`v_(y)=0-9.8xx10=98 m//s`
So speed `=98 sqrt(2) m//s`