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(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.Substituting the values: $T = \frac{2 	imes 20 	imes \sin(30^{\circ})}{10} = \

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$10$

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(A) The time of flight $T$ is given by $T = \frac{2u \sin 	heta}{g}$.Substituting the values: $T = \frac{2 	imes 20 	imes \sin(30^{\circ})}{10} = \frac{2 	imes 20 	imes 0.5}{10} = 2 \, s$.The time at which we need to find the centripetal acceleration is $t = 1 \, s$.Since the total time of flight is $2 \, s$,the object reaches its highest point at $t = \frac{T}{2} = 1 \, s$.At the highest point of a projectile's trajectory,the velocity is purely horizontal,and the acceleration is purely vertical (equal to $g$).The centripetal acceleration $a_c$ is defined as $a_c = \frac{v^2}{R}$. At the highest point,the radius of curvature $R$ is given by $R = \frac{v_x^2}{g}$.Substituting $R$ into the formula for $a_c$,we get $a_c = \frac{v_x^2}{(v_x^2/g)} = g$.Therefore,the centripetal acceleration at the highest point is $g = 10 \, m/s^2$.

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

An object is projected from the ground with a speed of $20 \, m/s$ at an angle of $30^{\circ}$ with the horizontal. Its centripetal acceleration one second after the projection is .......... $m/s^2$ [Take $g = 10 \, m/s^2$].

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