A ball is dropped from a height H from rest. | vedclass

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(B) Let $T$ be the total time taken to reach the ground. Then $H = \frac{1}{2} g T^2$.In the last $1.0 \ s$,the ball travels $\frac{H}{2}$. This means

Vedclass · Accepted Answer

$3.41$

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(B) Let $T$ be the total time taken to reach the ground. Then $H = \frac{1}{2} g T^2$.In the last $1.0 \ s$,the ball travels $\frac{H}{2}$. This means in the first $(T - 1) \ s$,the ball travels $\frac{H}{2}$.So,$\frac{H}{2} = \frac{1}{2} g (T - 1)^2$.Substituting $H = \frac{1}{2} g T^2$ into the equation:$\frac{1}{2} (\frac{1}{2} g T^2) = \frac{1}{2} g (T - 1)^2$$\frac{T^2}{4} = (T - 1)^2$Taking the square root on both sides:$\frac{T}{2} = T - 1$ (taking the positive root as $T > 1$)$T = 2T - 2$$T = 2 \ s$ (This is incorrect based on the quadratic solution).Let's re-solve: $\frac{T^2}{4} = T^2 - 2T + 1$$T^2 = 4T^2 - 8T + 4$$3T^2 - 8T + 4 = 0$Using the quadratic formula $T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$T = \frac{8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}$$T = \frac{12}{6} = 2 \ s$ or $T = \frac{4}{6} = 0.66 \ s$.Wait,the distance covered in the last second is $\frac{H}{2}$.$H - \frac{H}{2} = \frac{1}{2} g (T-1)^2 \implies \frac{1}{2} g T^2 - \frac{1}{2} g T^2 / 2 = \frac{1}{2} g (T-1)^2$ is wrong.Correct approach: Distance in last $1 \ s$ is $H - H_{T-1} = \frac{H}{2}$.$H_{T-1} = \frac{1}{2} g (T-1)^2 = \frac{H}{2} = \frac{1}{4} g T^2$.$(T-1)^2 = \frac{T^2}{2} \implies T^2 - 2T + 1 = \frac{T^2}{2} \implies \frac{T^2}{2} - 2T + 1 = 0 \implies T^2 - 4T + 2 = 0$.$T = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}$.Since $T > 1$,$T = 2 + 1.414 = 3.414 \ s$.

$A$ ball is dropped from a height $H$ from rest. The ball travels $\frac{H}{2}$ in the last $1.0 \ s$. The total time taken by the ball to hit the ground is: (in $s$)

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