From a tower of height h,one ball is thrown d | vedclass

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(C) For the first ball thrown vertically downwards:Using the equation of motion $v^2 = u^2 + 2as$,where $a = g$ and $s = h$,we get:$v_1^2 = u^2 + 2gh$

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$1 : 1$

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(C) For the first ball thrown vertically downwards:Using the equation of motion $v^2 = u^2 + 2as$,where $a = g$ and $s = h$,we get:$v_1^2 = u^2 + 2gh$$v_1 = \sqrt{u^2 + 2gh}$For the second ball thrown horizontally:The horizontal component of velocity remains constant: $v_x = u$.The vertical component of velocity when it reaches the ground is given by $v_y^2 = 0^2 + 2gh$,so $v_y = \sqrt{2gh}$.The resultant velocity $v_2$ is given by:$v_2 = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + (\sqrt{2gh})^2} = \sqrt{u^2 + 2gh}$Comparing the two velocities:$v_1 = v_2 = \sqrt{u^2 + 2gh}$Therefore,the ratio of their velocities is $v_1 : v_2 = 1 : 1$.

From a tower of height $h$,one ball is thrown downwards with a velocity $u$ and another ball is thrown horizontally with the same velocity $u$. What is the ratio of their velocities when they reach the ground?

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