A ball at point 'O' is at a horizontal distan | vedclass

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(B) The horizontal distance to the wall is $x = 7 \ m$. The initial velocity is $u = 15 \ m/s$ at an angle $	heta = 37^{\circ}$.First,calculate the h

Vedclass · Accepted Answer

$1.7$

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(B) The horizontal distance to the wall is $x = 7 \ m$. The initial velocity is $u = 15 \ m/s$ at an angle $	heta = 37^{\circ}$.First,calculate the height of the target '$C$' from the ground level '$B$':$	an 37^{\circ} = \frac{BC}{x} \implies BC = 7 	imes 	an 37^{\circ} = 7 	imes \frac{3}{4} = 5.25 \ m$.Next,find the time taken to reach the wall:$v_x = u \cos 37^{\circ} = 15 	imes \frac{4}{5} = 12 \ m/s$.$x = v_x \cdot t \implies 7 = 12 \cdot t \implies t = \frac{7}{12} \ s$.Now,calculate the vertical height '$BD$' reached by the ball at time '$t$':$v_y = u \sin 37^{\circ} = 15 	imes \frac{3}{5} = 9 \ m/s$.$BD = v_y \cdot t - \frac{1}{2} g t^2 = 9 	imes \left(\frac{7}{12}ight) - \frac{1}{2} 	imes 10 	imes \left(\frac{7}{12}ight)^2$.$BD = 5.25 - 5 	imes \frac{49}{144} = 5.25 - \frac{245}{144} \approx 5.25 - 1.701 = 3.549 \ m$.The vertical distance '$y_0$' is the difference between '$BC$' and '$BD$':$y_0 = BC - BD = 5.25 - 3.549 = 1.701 \ m \approx 1.7 \ m$.

$A$. 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 $ms^{-1}$)	$IV$. $22.4$

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