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(B) Let $t_A = 2 \ s$ be the time interval for boy $A$ and $t_B = 1 \ s$ be the time interval for boy $B$.For a ball projected vertically, the time ta

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(B) Let $t_A = 2 \ s$ be the time interval for boy $A$ and $t_B = 1 \ s$ be the time interval for boy $B$.For a ball projected vertically, the time taken to reach the maximum height from a point at height $h$ is $t/2$.Using $v = u + at$, at the highest point $v = 0$, so $u = g(t/2)$.The height $h$ of the point from the ground is given by $h = u(t/2) - 1/2 g(t/2)^2 = g(t/2)^2 - 1/2 g(t/2)^2 = 1/2 g(t/2)^2 = 1/8 g t^2$.For boy $A$: $h_A = 1/8 	imes 10 	imes (2)^2 = 5 \ m$.For boy $B$: $h_B = 1/8 	imes 10 	imes (1)^2 = 1.25 \ m$.The difference in vertical positions is $h_A - h_B = 5 \ m - 1.25 \ m = 3.75 \ m$.

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