A bottle dropped from a balloon reaches the ground in 20 s. determine the height of the balloon if (a) it was at rest in the air and (b) it was ascending with a speed of 50 m/s when the bottle was dropped.

a) 1960 m b) 960 m Assumptions. 1. Ignore air resistance. 2. Gravity is 9.80 m/s^2 For the situation where the balloon was stationary, the equation for the distance the bottle fell is d = 1/2 AT^2 d = 1/2 9.80 m/s^2 (20s) ^2 d = 4.9 m/s^2 * 400 s^2 d = 4.9 * 400 m d = 1960 m For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes d = 1/2 9.80 m/s^2 (20s) ^2 - 50 m/s * T d = 1/2 9.80 m/s^2 (20s) ^2 - 50m/s * 20s d = 4.9 m/s^2 * 400 s^2 - 1000 m d = 4.9 * 400 m - 1000 m d = 1960 m - 1000 m d = 960 m