A hiker stands on the edge of cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 "m s "^(-1) Neglecting airresistance , find the time taken by the stone to reach the ground , and the speed with which it hits the ground . (Take g=9.8 " m s"^(2)).

A hiker stands on the edge of cliff 490 m above the ground and throws

1.	A hiker stands on the edge of cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 "m s "^(-1) Neglecting airresistance , find the time taken by the stone to reach the ground , and the speed with which it hits the ground . (Take g=9.8 " m s"^(2)).
Answer» Solution :We choose the ORIGIN of 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 positivedirection of axis to be the VERTICALLY upward direction The x- , and y- COMPONENTS of the motion can be treated independently . Theequations of motion are : `x(t)=x_(@)+v_(@x)t` `y(t)=y_(@)+v_(@y)t+(1//2)a_(y)t^(2)` Here , `x_(@)=y_(@)=0,v_(@y)=0,a_(y)=-r=-9.8 " m s "^(-2),v_(@x)=15 " m s"^(-1)`. The stone hits the ground when y(t) =- 490 m. `-490m=-(1//2)(9.8)t^(2)`. This gives `t=10s`. The velocity components are `v_(x)=v_(@x)andv_(y)=v_(@y)- gt` so that when the stonehits the ground : `v_(@y)=15 " m s"^(-1)` `v_(@y)=0-9.8xx10=-98 " m s"^(-1)` Therefore , the speed of the stone is `sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(15^(2)+98^(2))=99 " m s"^(-1)`