A stone is thrown vertically up from the top | vedclass

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(D) Let the top of the window be the origin $(y=0)$. The stone is thrown upwards with $u = 8 \,m/s$. At the maximum height $A$, the final velocity $v

Vedclass · Accepted Answer

$0.2$

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(D) Let the top of the window be the origin $(y=0)$. The stone is thrown upwards with $u = 8 \,m/s$. At the maximum height $A$, the final velocity $v = 0$. Using $v = u + at$: $0 = 8 - 10t \implies t = 0.8 \,s$. The height of point $A$ above the top of the window is $h = ut + \frac{1}{2}at^2 = 8(0.8) - \frac{1}{2}(10)(0.8)^2 = 6.4 - 3.2 = 3.2 \,m$. Now, the stone falls from point $A$ to the bottom of the window. The total distance from $A$ to the bottom of the window is $H = 3.2 \,m + 1.8 \,m = 5.0 \,m$. Time taken to fall from $A$ to the bottom of the window $(t_{total})$: $H = \frac{1}{2}gt_{total}^2 \implies 5.0 = \frac{1}{2}(10)t_{total}^2 \implies t_{total}^2 = 1 \implies t_{total} = 1 \,s$. The time taken to cross the window during the downward journey is the difference between the time to reach the bottom and the time to reach the top of the window from $A$: $\Delta t = t_{total} - t = 1 \,s - 0.8 \,s = 0.2 \,s$.

$A$ stone is thrown vertically up from the top end of a window of height $1.8 \,m$ with a velocity of $8 \,m/s$. The time taken by the stone to cross the window during its downward journey is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,s$)

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