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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» <html><body><p></p>Solution :We choose the <a href="https://interviewquestions.tuteehub.com/tag/origin-1139399" style="font-weight:bold;" target="_blank" title="Click to know more about ORIGIN">ORIGIN</a> of the x - and y-axis at the <a href="https://interviewquestions.tuteehub.com/tag/edge-965664" style="font-weight:bold;" target="_blank" title="Click to know more about EDGE">EDGE</a> of the cliff and t =0 s at the instant the stone is <a href="https://interviewquestions.tuteehub.com/tag/thrown-7258593" style="font-weight:bold;" target="_blank" title="Click to know more about THROWN">THROWN</a> . Choose the positive direction of x-axis to be along the initial velocity and the positivedirection of axis to be the <a href="https://interviewquestions.tuteehub.com/tag/vertically-3260386" style="font-weight:bold;" target="_blank" title="Click to know more about VERTICALLY">VERTICALLY</a> upward direction The x- , and y- <a href="https://interviewquestions.tuteehub.com/tag/components-926700" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENTS">COMPONENTS</a> of the motion can be treated independently . Theequations of motion are :<br/>`x(t)=x_(@)+v_(@x)t`<br/>`y(t)=y_(@)+v_(@y)t+(1//2)a_(y)t^(2)`<br/>Here , `x_(@)=y_(@)=0,v_(@y)=0,a_(y)=-r=-9.8 " m s "^(-2),v_(@x)=15 " m s"^(-1)`.<br/>The stone hits the ground when y(t) =- 490 m. <br/>`-490m=-(1//2)(9.8)t^(2)`.<br/>This gives `t=10s`.<br/>The velocity components are `v_(x)=v_(@x)andv_(y)=v_(@y)- gt`<br/>so that when the stonehits the ground :<br/>`v_(@y)=15 " m s"^(-1)`<br/>`v_(@y)=0-9.8xx10=-98 " m s"^(-1)`<br/>Therefore , the speed of the stone is<br/>`sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(15^(2)+98^(2))=99 " m s"^(-1)`</body></html>


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