A hot-air balloon is rising straight upward with a constant of 6.5 m/s. When the basket of the balloon is 20.0 meters above the ground a bag of sand comes loose. How long does it take the bag of sand to hit the ground? What is the greatest height of the bag during its fall to the ground?

T = 2.79 sec.

s = 2.16 m

Explanation:

from free fall formula we have following relation

S = VoT + (1/2) gT^2

where

S = distance from ground which is given 20 meters

Vo = initial velocity = - 6.5 m/sec

T = time for bag to hit the ground

putting all appropriate values,

20 = - (6.5T) + (1/2) (9.8) T^2

solving for T

4.9T^2 - 6.5T - 20 = 0

T = 2.79 sec.

B)

V_f^2 - V_o^2 = 2gs

where

V_f = final velocity = 0

V_o = - 6.5 m/sec

g = 9.8 m/sec^2

s = height of bag before it strart to hit the ground

putting values,

0 - (-6.5) ^2 = 2 (9.8) s

42.25 = 19.6 (s)

solving for "s",

s = 42.25 / (2 * 9.8)

s = 2.16 m

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