A body is projected horizontally with a velocity of `10ms^(-1)` from the top of building 20 m high. Find
(a) horizontal distance from the bottom of the building at which the body will strike the ground.
(b) the magnitude and direction of the velocity of the body 1s after it is projected. Take `g=10ms^(-2)`.

Explanation

(a) Refer figure.

Given `h=-20m," "u=10ms^(-1), and g=-10ms^(-2)`. Time taken by the body to go from A to B is
`t=sqrt((2h)/(g))=sqrt((2xx-20)/(-10))=2s`
Horizontal distance OR=R=ut=`10xx2=20`m
(b) Refer to figure again. horizontal velocity at `t=1s` is `upsilon_(x)=u=10ms^(-1)`. vertical velocity at t=1 s is (since initial vertical component of velocity `u_(y)=0`)
`upsilon_(y)=u_(y)+at=0-10xx1 implies upsilon_(y)=-10ms^(-1)`
Magnitude of resultant velocity is `upsilon=sqrt(upsilon_(x)^(2)+upsilon_(y)^(2))=sqrt(10^(2)+(-10)^(2))`
`=sqrt(200)=10sqrt(2)ms^(-1)`
The angle `alpha` which the resultant velocity vector subtends with the vertical is given by
`tan alpha=(upsilon_(x))/(|upsilon_(y)|)=(10)/(10)=1 implies alpha=45^(@)`.