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Step-by-Step Solution
Step 1: Identify known quantities and unknowns
• Total height of the tower, H = 100\,\text{m} .
• Distance covered during the last 0.5\,\text{s} of the motion is 19\,\text{m} .
• Let the acceleration due to gravity on that planet be a \,(\text{m/s}^2) .
• Let t be the time needed for the ball to travel the first 81\,\text{m} (because 100 - 19 = 81 ). Then the total time to travel 100\,\text{m} is (t + 0.5)\,\text{s} .
Step 2: Write down the equation for distance traveled in time t
Since the ball is dropped (initial velocity u = 0 ) and moves under constant acceleration a , the distance s traveled in time t is:
s = \frac{1}{2} a t^2.
For the first 81\,\text{m} :
81 = \frac{1}{2}\,a\,t^2.
So,
t^2 = \frac{162}{a} \quad\Longrightarrow\quad t = \sqrt{\frac{162}{a}} = 9\,\sqrt{\frac{2}{a}}.
Step 3: Write the equation for total distance ( 100\,\text{m} ) in (t + 0.5)\,\text{s}
Similarly, for the total distance of 100\,\text{m} covered in (t + 0.5)\,\text{s} :
100 = \frac{1}{2}\,a\,(t + 0.5)^2.
Taking the square root on both sides (when rearranged):
t + 0.5 = \sqrt{\frac{200}{a}} = 10\,\sqrt{\frac{2}{a}}.
Step 4: Relate the two times to solve for a
From Step 2, we have t = 9\,\sqrt{\frac{2}{a}}. Substituting this into t + 0.5 = 10\,\sqrt{\frac{2}{a}} :
9\,\sqrt{\frac{2}{a}} + 0.5 = 10\,\sqrt{\frac{2}{a}}.
Rearranging,
10\,\sqrt{\frac{2}{a}} - 9\,\sqrt{\frac{2}{a}} = 0.5 \quad\Longrightarrow\quad \sqrt{\frac{2}{a}} = 0.5.
Therefore,
\frac{2}{a} = 0.5^2 = 0.25 \quad\Longrightarrow\quad a = \frac{2}{0.25} = 8\,\text{m/s}^2.
Step 5: Conclude the value of the planetary acceleration
Hence, the acceleration due to gravity on that planet is:
\boxed{8\,\text{m/s}^2}.
