A plane flies from base camp to lake a, a distance of 290 km at a direction of 20.0° north of east. after dropping off supplies,
1 answer:
We can simply find components of the given directions and conveniently add them up to find the total distance.
290 km 20 deg north of east can be decomposed as :
290 x cos(20) EAST + 290 sin(20) NORTH
=272.5 E + 99.18 N
Now from this point ,
170 Km 30 deg west of north :
170 cos(30) N - 170 sin (30) E [ east = - west]
= 147.22 N - 85 E
hence coordinates of lake b are
(272.5 - 85) E + (99.18 + 147.22)N
187.5 E +264.4 N
use right triangle property to find distance D =
D = 324.13 Km
and angle = atan(b/a) = 54.65 deg
hence lake b is at 324.13 Km 54.65 deg east of north
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Explanation:
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where:
is the initial potential energy, at the bottom
is the initial kinetic energy, at the bottom
is the final potential energy, at the top
is the final kinetic energy, at the top
We can rewrite the equation as:
where:
m is the mass of the hopper
is the acceleration of gravity
is the initial height
u is the launch speed of the hopper
is the maximum altitude reached by the hopper
v = 0 is the final speed (which is zero when the hopper reaches the maximum height)
Solving the equation for u, we find the launch speed of the hopper:
Learn more about kinetic energy and potential energy:
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