A hiker stands on the edge of a cliff `490 m` above the ground and throwns a stone horiozontally with an initial speed of `15ms^(-1)` neglecting air resistance.The time taken by the stone to reach the ground in seconds is `(g=9.8ms^(2))`

Explanation

We choosethe origin at the x-, and y-axis at the edge of the cliff and t=0 s at the instant the stone is thrown. Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to bethe vertically upward direction. The x-and y- componetns of the motion can be treated independently. The equation of motion are:
`x(t) = x_(0) + v_(ox)t`
`y(t) =y_(0) + v_(oy) t+(1//2)a_(y)t^(2)`
Here, `x_(0) = y_(0)=0, v_(oy)=0, a_(y)=-g=-9.8ms^(-2)`.
`v_(ox) = 15ms^(-1)`.
The stone hits hte ground when y(t) = `-490m`.
`-490m = -(1//2)(9.8)t^(2)`.
This gives, `t=10s`
The velocity componetns are `v_(x) =v_(ox)` and `v_(y) = v_(oy)-gt`
So that when the stone hits the ground:
`v_(ox) = 15ms^(-1)`
`v_(oy)=0-9.8 xx 10= -9.8ms^(-1)`
Therefore, the speed of the stone is
`sqrt(v_(x)^(2)+v_(y)^(2)) = sqrt(15^(2)+98^(2))` = 99`ms^(-1)`.