hmm..
Vx= 220cos30 = 190.5
Vy = 110
s=-Vy*t + (1/2)*a*t^2
2800 = -110*t + (1/2)*(9.81)*t^2
use quadratic equation to solve for t
t=37.6s
v=g*t
v=9.81 * 37.6
v=368.85
tan^-1(v/vx)
tan^-1(368.85/110) = 73.39
See if that's correct..
An airplane is flying with a velocity of 220 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.8 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle θ?
I found
Vox=190.53
Voy=110
T=110/9.8=11.22
X=(190.53)(22.44)=4277.4
y=(110)(11.22)+.5(9.8)(11.22)^2
y=617.339
4277.2^2+617.339^2=c^2
C=4321.72
inv_sin(617.339/4321.72)=8.21
But my answer wasn't correct, What am I doing wrong?
2 answers
oops
tan^-1(368.85/190.5) = 62.68
tan^-1(368.85/190.5) = 62.68