A hiker stands on the edge of a cliff $490\; m$ above the ground and throws a stone horizontally with an initial speed of $15 \;m/ s$. Neglecting air resistance, 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 a cliff $490\; m$ above the ground and throws a stone horizontally with an initial speed of $15 \;m/ s$. Neglecting air resistance, 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 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 positive direction of $y$ -axis to be the vertically upward direction. The $x-,$ and $y$ components of the motion can be treated independently. The equations of motion are:
$x(t) =x_{o}+v_{\alpha x} t$
$y(t) =y_{o}+v_{o y} t+(1 / 2) a_{y} t^{2}$
Here, $x_{0} =y_{0}=0, v_{o y}=0, a_{y}=-g=-9.8 m s ^{-2}$
$v_{ox }=15 m s ^{-1}$
The stone hits the ground when $y(t)=-490 m$
$-490 m =-(1 / 2)(9.8) t^{2}$
This gives $t=10 s$ The velocity components are $v_{x}=v_{a x}$ and $v_{y}=v_{o y}-g t$
so that when the stone hits the ground:
$v_{o x}=15 m s ^{-1}$
$v_{o y}=0-9.8 \times 10=-98 m s ^{-1}$
Therefore, the speed of the stone is
$\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{15^{2}+98^{2}}=99 m s ^{-1}$
Similar Questions
The position vector of an object at any time $t$ is given by $3 t^2 \hat{i}+6 t \hat{j}+\hat{k}$. Its velocity along $y$-axis has the magnitude
The position vector of an object at any time $t$ is given by $3 t^2 \hat{i}+6 t \hat{j}+\hat{k}$. Its velocity along $y$-axis has the magnitude
A particle projected from origin moves in $x-y$ plane with a velocity $\vec{v}=3 \hat{i}+6 x \hat{j}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. Find the equation of path followed by the particle
A particle projected from origin moves in $x-y$ plane with a velocity $\vec{v}=3 \hat{i}+6 x \hat{j}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along $x$ and $y$ axis. Find the equation of path followed by the particle
The coordinates of a particle moving in a plane are given by $x = a\cos (pt)$ and $y(t) = b\sin (pt)$ where $a,\,\,b\,( < a)$ and $p$ are positive constants of appropriate dimensions. Then
The coordinates of a particle moving in a plane are given by $x = a\cos (pt)$ and $y(t) = b\sin (pt)$ where $a,\,\,b\,( < a)$ and $p$ are positive constants of appropriate dimensions. Then
- [IIT 1999]
A river is flowing due east with a speed $3\, ms^{-1}$. A swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction) ?
$(b)$ If he wants to start from point A on south bank and reach opposite point $B$ on north bank,
$(i)$ Which direction should he swim ?
$(ii)$ What will be his resultant speed ?
$(c)$ From two different cases as mentioned in $(a)$ and $(b)$ above, in which case will he reach opposite bank in shorter time ?
A river is flowing due east with a speed $3\, ms^{-1}$. A swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction) ?
$(b)$ If he wants to start from point A on south bank and reach opposite point $B$ on north bank,
$(i)$ Which direction should he swim ?
$(ii)$ What will be his resultant speed ?
$(c)$ From two different cases as mentioned in $(a)$ and $(b)$ above, in which case will he reach opposite bank in shorter time ?
Give explanation of position and displacement vectors for particle moving in a plane by giving suitable equations.
Give explanation of position and displacement vectors for particle moving in a plane by giving suitable equations.