A hiker stands on the edge of a cliff 490 ; m | vedclass

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(D) 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. The positive direction

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$10 \; s$ and $99 \; m/s$

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(D) 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. The positive direction of the $x$-axis is along the initial velocity,and the positive direction of the $y$-axis is vertically upward.The motion in $x$ and $y$ directions can be treated independently. The equations of motion are:$x(t) = x_0 + v_{0x} t$$y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2$Given: $x_0 = 0, y_0 = 0, v_{0x} = 15 \; m/s, v_{0y} = 0, a_y = -g = -9.8 \; m/s^2$.The stone hits the ground when $y(t) = -490 \; m$:$-490 = 0 + 0(t) + \frac{1}{2}(-9.8)t^2$$-490 = -4.9 t^2$$t^2 = 100 \implies t = 10 \; s$.The velocity components at time $t = 10 \; s$ are:$v_x = v_{0x} = 15 \; m/s$$v_y = v_{0y} - gt = 0 - 9.8(10) = -98 \; m/s$The speed of the stone is the magnitude of the velocity vector:$v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + (-98)^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99.14 \; m/s$.Rounding to the nearest integer,the speed is $99 \; m/s$.

List-$I$ (Kinematic Variable)	List-$II$ (Value)
$A$. Velocity of the body after $1 \ s$ (in $ms^{-1}$)	$I$. $5$
$B$. Horizontal displacement of the body after $1 \ s$ (in $m$)	$II$. $20$
$C$. Vertical displacement of the body after $1 \ s$ (in $m$)	$III$. $10$
$D$. Vertical velocity of the body after $1 \ s$ (in $ms^{-1}$)	$IV$. $22.4$

$A$ 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 $g = 9.8 \; m/s^2$).

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