Question

A hiker stands on the edge of a cliff 490 , above the ground and throws a stone horizontally with an initial speed of $15m{s}^{{}_1}$. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. $(Takeg=9.8m{s}^{-2})$.

Answer 1

motion of stone may be considered as the super position of the two independent sources given horizontal with constant velocity $u=15m/s$

vertices motion with constant acceleration $a=g=9.8m/s^2$

Let $h$ be the height of the cliff above ground

Let $u_v$ be the vertical component of velocity of projection of the stone If the stone hits the ground after $1$ second of projection the $h=u_{v}t+\frac{1}{2}g{t}^2$

the stone is thrown horizontally vertical component

$\therefore h=0+\frac{1}{2}g{t}^2\Rightarrow t=\frac{\sqrt{2h}}{9}$

This gives time for stone to reach ground

$t=\frac{\sqrt{2h}9}{=}\sqrt{\frac{2\times 490}{9.8}}=\sqrt{\frac{980}{9.8}}=\sqrt{100}=10\Delta$

$t=10\Delta$

Let $V_4$ be the vertical component of velocity of stone when it hots the ground then

$v_y^2\pm u_y^2+2gh$

$v_y^2=(0{)}^2+2\times +9.8\times 490$

$v_y^2=9604\Rightarrow V_y=\sqrt{9604}$

$=98m/s$

The horizontal component of velocity $V_x$ with which the hits teh ground it remain constant because there is no acceleration is horizontal direction

$\therefore V_x=u_x=15m/s$

$\therefore$ final speed of stone

$V=\sqrt{V_x^2+V_y^2}=\sqrt{{15}^2}+{98}^2$

$=\sqrt{9829}=99.14m/s$

$\therefore$ speed with which the stone hits the ground is $99.4m/s$