Let the particle be fired from the top (O) of a hill of heith `h= AO` =500 m` velocity `u=100 ms^(-1)` along the horizontal direction. It reaches the ground at (B) after time (t).
(i) Taking vertical downward motion of the particle from (O) to (B), we have,
`y_(0) =0, y=500 m, u_(y) =0, a_(y) =10 ms^(2), t=?`
As `y= y_(0) + u_(y) t + 1/2 a_(y) t^(2)`
:. `500 =0 +o xx t + 1/2 10 xx t^(2)`
or `t^(2) =100 or `t=10 s.`
.
(ii) Distance of the target from the hill =`AB=x`
Taking horizontal motion of particle from (O) to (B), we have
`x_(0) =0, z=?, u_(x) =100 ms^(-1)` a_(x) =0, t=10 s`
As, `x=x_(0) +u_(x) t+ 1/2 a_(x) t^(2)`
:. `x=0 + 100 xx 10 + 1/2 xx 0 xx 10^(2) =1000 m`
(ii) Let `v_(x) and v_(y)` be the borizontal and vertical components of velocity `v` of the particle at (B).
Then `v_(x) =u_(x) a_(x) t=u_(x) =100 ms^(-1)`
`v_(y) =u_(y) + a_(y) t=0 + 10 xx 10 =100 ms^(-1)`
`v= sqrt (v_(x)^(2) +v_(y)^(2)) =sqrt( 100^(2) + 100^(2)) = 100 sqrt 2 ms^(-1)`
Let the resultant velocity at (B) maken an angle `brta` with the horizontal direction , then
`tan beta =v_(y)/v_(x) =(100)/(100) =1=tan 45^(@) or beta =45^(@)`.
