A 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. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.
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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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