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 wil strike the ground.
b. The magnitude and direction of the velocity of the 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((2x-20)/(-10))=2s`
Horizontal disance `OB=R=ut=10xx2=20m`
b. Refer to figure agai. Horizontal velocity at t=1s is `v_(x)=u=10ms^(-1)`. Vertical velocity at t=1s is (since initial vertical component of velocity `u_(u)=0`)
`v_(y)=u_(y)+at=0-10xx1impliesv_(y)=-10ms^(-1)`
Magnitude of resultant velocity is `v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(10^(2)+(-10)^(20)=sqrt(200)=10sqrt(2)ms^(-1)`
The angle `alpha` which the resultant velocity vector subtends with the vertical is given by
`tan alpha=(v_(x))/(|v_(y)|)=10/10=1impliesalpha=45^(@)`