(A) Given:Initial velocity,$u = 20\; m/s$Final velocity,$v = 80\; m/s$Acceleration due to gravity,$g = 10\; m/s^2$Using the third equation of motion:$

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(A) Given:Initial velocity,$u = 20\; m/s$Final velocity,$v = 80\; m/s$Acceleration due to gravity,$g = 10\; m/s^2$Using the third equation of motion:$v^2 = u^2 + 2gh$Substituting the values:$80^2 = 20^2 + 2 \times 10 \times h$$6400 = 400 + 20h$$6000 = 20h$$h = \frac{6000}{20} = 300\; m$Therefore,the height of the tower is $300\; m$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

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$A$ ball is thrown vertically downward with a velocity of $20\; m/s$ from the top of a tower. It hits the ground after some time with a velocity of $80\; m/s$. The height of the tower is $......m$ : $(g = 10\; m/s^2)$

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Motion in Straight Line

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Motion Under Gravity

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$A$ ball is thrown vertically downward with a velocity of $20\; m/s$ from the top of a tower. It hits the ground after some time with a velocity of $80\; m/s$. The height of the tower is $......m$ : $(g = 10\; m/s^2)$

Similar Questions

$A$ body falling from a vertical height of $10 \, m$ pierces through a distance of $1 \, m$ in sand. It faces an average retardation in sand equal to (where $g$ = acceleration due to gravity).

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