Nice question.
One has to assume that the angle of elevation of 30º does not change in the short interval of our event.
Second assumption: the sandbag falls vertically, without forward motion.
At a time of t seconds, let the shadow of the sandbag be x m from the vertical, and let s be height of the sandbag.
so tan 30 = s/x
s = √3x
ds/dt = √3dx/dt
but ds/dt = -9.8t
when s = 35
35 = 60 - 4.9t^2
4.9t^2 = 25
t = √(25/4.9) = 2.259 sec
then ds/dt = -9.8(2.259) m/s
= -22.136 m/s
so dx/dt = (ds/dt)/√3
= -22.136/√3
= -12.78 m/s
a sandbag is dropped from a balloon at a height of 60 meteres when the angle of eleveation to the sun is 30 degrees. find the rate at which the shadow of the sandbag is traveling along the ground when the sandbag is at a height of 35 meters. hint: the position of the sandbag is give by s(t)=60-4.9t^2
how would you set up to solve this problem?
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