A cricketer hits a ball with a velocity 25,m/ | vedclass

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(A) The horizontal component of the initial velocity is $v_x = v \cos 	heta = 25 \cos 60^\circ = 25 	imes 0.5 = 12.5\,m/s$.The vertical component of

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$8.2$

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(A) The horizontal component of the initial velocity is $v_x = v \cos 	heta = 25 \cos 60^\circ = 25 	imes 0.5 = 12.5\,m/s$.The vertical component of the initial velocity is $v_y = v \sin 	heta = 25 \sin 60^\circ = 25 	imes \frac{\sqrt{3}}{2} = 12.5\sqrt{3}\,m/s$.The time $t$ taken to cover a horizontal distance of $x = 50\,m$ is $t = \frac{x}{v_x} = \frac{50}{12.5} = 4\,s$.The vertical height $y$ at time $t$ is given by the equation of motion $y = v_y t - \frac{1}{2} g t^2$.Substituting the values: $y = (12.5\sqrt{3}) 	imes 4 - \frac{1}{2} 	imes 9.8 	imes (4)^2$.$y = 50\sqrt{3} - 4.9 	imes 16 = 50 	imes 1.732 - 78.4 = 86.6 - 78.4 = 8.2\,m$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

$A$ cricketer hits a ball with a velocity $25\,m/s$ at $60^\circ$ above the horizontal. How far above the ground does it pass over a fielder $50\,m$ from the bat (in $,m$)? (Assume the ball is struck very close to the ground and $g = 9.8\,m/s^2$)

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Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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