(C) Let $u$ be the velocity of the ball at the top of the window.Using the second equation of motion for the motion of the ball across the window:$S =

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(C) Let $u$ be the velocity of the ball at the top of the window.Using the second equation of motion for the motion of the ball across the window:$S = ut + \frac{1}{2}at^2$Here,$S = 1.5 \; m$,$t = 0.1 \; s$,and $a = g = 10 \; m/s^2$.Substituting the values:$1.5 = u(0.1) + \frac{1}{2} \times 10 \times (0.1)^2$$1.5 = 0.1u + 5 \times 0.01$$1.5 = 0.1u + 0.05$$0.1u = 1.5 - 0.05$$0.1u = 1.45$$u = \frac{1.45}{0.1} = 14.5 \; m/s$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

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$A$ person sitting on the ground floor of a building notices through a window of height $1.5 \; m$ that a ball dropped from the roof of the building crosses the window in $0.1 \; s$. What is the velocity (in $m/s$) of the ball when it is at the topmost point of the window? (Take $g = 10 \; m/s^2$)

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$A$ person sitting on the ground floor of a building notices through a window of height $1.5 \; m$ that a ball dropped from the roof of the building crosses the window in $0.1 \; s$. What is the velocity (in $m/s$) of the ball when it is at the topmost point of the window? (Take $g = 10 \; m/s^2$)

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