Youngkwon C. answered 03/12/16
Tutor
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Knowledgeable and patient tutor with a Ph.D. in Electrical Eng.
Hi Parveen,
According to the trigonometry of a triangle,
the shadow length, l(t), on the ground can be obtained as
l(t) = s(t)/tan(30º)
= √3(60 - 4.9t2)
According to the trigonometry of a triangle,
the shadow length, l(t), on the ground can be obtained as
l(t) = s(t)/tan(30º)
= √3(60 - 4.9t2)
And, the shadow's travelling rate on the ground
is given by the first derivative of l(t),
l'(t) = -9.8√3t
is given by the first derivative of l(t),
l'(t) = -9.8√3t
The time when the bag was 35 meters above the ground
can be obtained by solving
60 - 4.9t2 = 35
t = 5√10/7
can be obtained by solving
60 - 4.9t2 = 35
t = 5√10/7
Plugging t = 5√10/7 into l'(t),
l'(5√10/7) = -9.8√3(5√10/7)
= -7√30
= -38.34 [m/s]
l'(5√10/7) = -9.8√3(5√10/7)
= -7√30
= -38.34 [m/s]
Parveen S.
03/13/16