The ball was thrown vertically upwards from a balcony at a height of 12 m.
| March 31, 2021 | education
The ball was thrown vertically upwards from a balcony at a height of 12 m. After 4 seconds the ball fell to the ground. Determine the initial speed.
Data: h1 (balcony height) = 12 m; t (duration of ball movement) = 4 s.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
Movement time: t = t1 + t2 = V0 / g + √ (2 * (h1 + h2) / g) = V0 / g + √ (2 * (h1 + V0 ^ 2 / 2g) / g).
Substitute the known values: 4 = V0 / 10 + √ (2 * (12 + V0 ^ 2 / (2 * 10)) / 10).
4 = V0 / 10 + √ ((12 + V0 ^ 2/20) / 5).
4 = 0.1V0 + √ (2.4 + 0.01V0 ^ 2).
(4 – 0.1V0) ^ 2 = 2.4 + 0.01V0 ^ 2.
16 – 0.8V0 + 0.01V0 ^ 2 = 2.4 + 0.01V0 ^ 2.
0.8V0 = 16 – 2.4 = 13.6 and V0 = 13.6 / 0.8 = 17 m / s.
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