Motion in 2-D question?

Question

A small pane takes off at a constant velocity of 150 km/hour at an angle of 37 degrees. At 3.00 s a)how high is the plane above ground and b) what horizontal distance has the plane traveled from the liftoff point?

Answer 1

first get the speed in seconds
150 km/hour
3600 seconds in an hour>>150 km per hour/3600 seconds per hour=5/120 km per second
the distance the plane travels is the hypotenuse (h)
the horizontal distance traveled is the adjacent angle(a)
the height traveled is the opposite angle(o)
and the angle of travel (37 degrees)
the trick to this problem is the memory aid SOH-CAH-TOA
sin=opposite/hypotenuse
cos=adjacent/hypotenuse
tan=opposite/adjacent
hypotenuse=5/120 km per second*3 seconds=5/40 km or .125km
we know the hypotenuse and the angle so we'll use the first two equations but rearranged
A) opposite(this is the height)=sin*hypotenuse=sin(37)*.125 km=.0752 km
B) adjacent(this is the horizontal)=cos*hypotenuse=cos(37)*.125 km=.0998 km
this side could also be solved using Pythagorean theorem
A^2+B^2=C^2 with the opposite and adjacent angles being A & B and the hypotenuse being C

Answer 2

resolve it's constant velocity into a vector form. The x component is 150*cos(37) the y component is 150*sin(37)

consider the horizontal first

150cos(37)=119.8 km/hr. 3 seconds is 3/3600 hrs. so the plane has traveled 119.9 km/hr*3/3600 hrs = .1 km

consider the verticle

150 sin(37)=90.27. same calculation 90.27*3/3600=.07 km