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(A) Let the height of the tower be $h$. Using the equation of motion $s = ut + \frac{1}{2}at^2$ with downward direction as positive:For ball $B$ (thro

Vedclass · Accepted Answer

$t_A > t_B$

Vedclass · Answer

(A) Let the height of the tower be $h$. Using the equation of motion $s = ut + \frac{1}{2}at^2$ with downward direction as positive:For ball $B$ (thrown downwards): $h = ut_B + \frac{1}{2}gt_B^2$. This is a quadratic equation in $t_B$.For ball $A$ (thrown upwards): $-h = ut_A + \frac{1}{2}g(-t_A)^2$ is incorrect; rather,using displacement $s = -h$ and initial velocity $u_A = -u$ (upwards is negative),we get $-h = -ut_A + \frac{1}{2}gt_A^2$,which simplifies to $h = ut_A - \frac{1}{2}gt_A^2$.Alternatively,consider the energy conservation or velocity at the point of return: Ball $A$ will eventually pass the top of the tower with velocity $u$ downwards,at which point it is effectively in the same state as ball $B$ at the start. Thus,the time taken for ball $A$ to reach the ground is the time to return to the tower top plus the time taken by ball $B$. Therefore,$t_A > t_B$.

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