From the top of a 64, ft high tower,a stone i | vedclass

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(D) Initial velocity $u = 48\, ft/s$.Acceleration due to gravity $g = 32\, ft/s^2$ (acting downwards).At the maximum height,the final velocity $v = 0$

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$100$

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(D) Initial velocity $u = 48\, ft/s$.Acceleration due to gravity $g = 32\, ft/s^2$ (acting downwards).At the maximum height,the final velocity $v = 0$.Using the kinematic equation $v^2 = u^2 - 2gh$,where $h$ is the height reached above the point of projection:$0^2 = (48)^2 - 2(32)h$$0 = 2304 - 64h$$64h = 2304$$h = \frac{2304}{64} = 36\, ft$.The total height from the ground is the sum of the tower height and the height reached above the tower:$H_{max} = 64\, ft + 36\, ft = 100\, ft$.

From the top of a $64\, ft$ high tower,a stone is thrown upwards vertically with the velocity of $48\, ft/s$. The greatest height (in $ft$) attained by the stone,assuming the value of the gravitational acceleration $g = 32\, ft/s^2$,is

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