The discussion focuses on the physics of a package dropped from a hot-air balloon ascending at 12 m/s from a height of 80 m. The time taken for the package to reach the ground is calculated to be 5.4 seconds, with a final impact speed of 41.38 m/s. Key equations used include kinematic equations for uniformly accelerated motion, specifically the SUVAT equations. The initial upward velocity of the package is 12 m/s, and the acceleration due to gravity is -9.8 m/s².
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SteamKing said:I think here, instead of introducing spurious negative signs, it would be better to use g = -9.8 m/s2 along with the distance fallen of -87.347 m.
v02 still comes out positive.
SteamKing said:The change in position will also be negative since the package was 87.347 m above the ground before it fell.
Eclair_de_XII said:But won't it be imaginary since ##v_0## is still negative?
The equation is v2 = u2 + 2 a s
There is no negative sign on the velocity terms. The product of -9.8 and -87.347 is also positive.
But then you'd have to move the negative displacement to the other side, making it positive.
Not if you take x = 0 being the ground and x0 = 87.347 m as the height from which the package started falling. Here, v0 = 0 and g = -9.8 m/s2, making the equation:
x - x0 = v0 t + (1/2)*g*t2 → 0 - 87.347 = (1/2) * (-9.8) * t2