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(B) Let the points of motion be $A$ (jump), $B$ (parachute opens), $C$ (at $10 \ m$ height), and $D$ (ground).$1.$ Motion from $A$ to $B$ (Free fall):

Vedclass · Accepted Answer

$92.5$

Vedclass · Answer

(B) Let the points of motion be $A$ (jump), $B$ (parachute opens), $C$ (at $10 \ m$ height), and $D$ (ground).$1.$ Motion from $A$ to $B$ (Free fall):Initial velocity $u_A = 0$, time $t = 2 \ s$, acceleration $a = g = 10 \ m/s^2$.Distance covered $x_1 = \frac{1}{2} g t^2 = \frac{1}{2} 	imes 10 	imes (2)^2 = 20 \ m$.Velocity at $B$, $v_B = u_A + gt = 0 + 10 	imes 2 = 20 \ m/s$.$2.$ Motion from $B$ to $C$ (Deceleration):Initial velocity $u_B = 20 \ m/s$, final velocity $v_C = 5 \ m/s$, acceleration $a = -3 \ m/s^2$.Using $v^2 - u^2 = 2as$:$(5)^2 - (20)^2 = 2(-3) x_2$$25 - 400 = -6 x_2$$-375 = -6 x_2$$x_2 = \frac{375}{6} = 62.5 \ m$.$3.$ Motion from $C$ to $D$:Distance $x_3 = 10 \ m$.Total height $H = x_1 + x_2 + x_3 = 20 + 62.5 + 10 = 92.5 \ m$.

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