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let it takes t second to reach the top of the window.
So it will take (t+0.1) s to reach the bottom of the window.
distance between them = 2m
So
(0×t + 0.5×10×(t+0.1)²) - (0×t + 0.5×10×(t)²) = 2
⇒ 5(t+0.1)² - 5t² = 2
⇒ 5(t² + 0.01 + 0.2t) - 5t² = 2
⇒ 5t² + 0.05 + t - 5t² = 5
⇒ t + 0.05 = 2
⇒ t = 5 - 0.05 = 4.95s
Distance of top of window = 0.5×10×4.95² = 5×24.5025 = 122.5m
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