A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sandbag at the instant the balloon is 60.0 m above the ground. After it is released, the sandbag encounters no appreciable air resistance.
a) Find the position and velocity of the sandbag at 0.200 s and 0.800 s after its release.
b) How much time does it take for the sandbag to strike the ground after it is released?
(a)
Position of sandbag 0.2 s after its release,
y = yo + vo*t + (1/2)at^2
y = 60 + 5*0.2 - (1/2)*9.8*0.22
y = 60.804 m
Position 0.8 s after its release,
y = yo + vo*t + (1/2)at^2
y = 60 + 5*0.8 - (1/2)*9.8*0.82
y = 60.86 m
Velocity of sandbag after 0.2 s,
v = vo + at
v = 5 - 9.8*0.20
v = 3.04 m/s
Velocity after 0.8 s,
v = 5 - 9.8*0.80
v = -2.84 m/s
(b)
Consider upward direction positive and downward direction negative .
From kinematic equation,
h = vo*t + (1/2)at^2
60 = -5*t + (1/2)*9.8*t^2
4.9t^2 - 5t - 60 = 0
By solving above quadratic equation,
t = 4.046 s
A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sandbag at...
2) Full of hot air. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sandbag at the instant the balloon is 60.0 m above the ground. After it is released, the sandbag encounters no appreciable air resistance a) Find the position and velocity of the sandbag at 0.200 s and 0.800 s after its release. (6) How much time does it take for the sandbag to strike the ground after it is released? (4) b)
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