That\'s a lot of hot-air balloonist, rising vertically with a constant speed of
ID: 2163517 • Letter: T
Question
That's a lot of hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. (See Figure 2.52.) After it is released, the sandbag encounters no appreciable' air drag, Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release, How many seconds after its release will the bag strike the ground? How fast is it moving as it strikes the ground? What is the greatest height above the ground that the sandbag reaches? Sketch graphs of this bag s acceleration, velocity, and vertical position as function ofExplanation / Answer
1. here, u= -5, g=9.81, t=.262 s= ut + (1/2) gt^2 = -5*.262 + .5*9.81*.262^2 = -.9733 height of sandbag above ground at .262s is 40+.9733=40.9733 m 2. here, u =-5, t=.262, g=9.81, v=? v=u+gt = -5+9.81*.262 = -2.42978 m/s velocity of sandbag at .262 s after release is 2.42978 m/s upwards. 3. t=1.1 s v=u+gt = -5 + 9.81*1.10 = 5.791 velocity of sandbag at 1.10 s after release is 5.791 m/s downwards. 4. u=-5, h=40, g=9.81, t=? h= ut + .5 gt^2 40= -5t + .5*9.81 t^2 4.905t^2-5t-40=0 this quadratic equation gives two solutions t=3.41 and -2.39 accepting only positive value, t=3.41 s 5. u=-5, t=3.41s, g=9.81, v=? v=u+gt =-5 + 9.81*3.41 =28.4521 at the moment it strikes ground, v=28.4521 m/s 6. u=5, v=0, g=-9.81, h=?(in this case upward being positive) v^2 = u^2 + 2gh 0=5^2 - 2*9.81*h h=1.274m greatest height above ground 40+h=41.274m
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