An airplane is flying with a velocity of 252 m/s at an angle of 30.0° with the horizontal, as the drawing shows. when the altitude of the plane is 2.3 km, a flare is released from the plane. the flare hits the target on the ground. what is the angle θ?

answered θ=40.52°

Step-by-step explanation:

We know that

S= 2300m , u= 252 m/s, a=9.8 m/s^2

s= u×t +1/2at^2

putting values we get

-2300= -252sin30°×t+ 0.5(-9.8)t^2

4.9t^2+126t-2300=0

⇒t= 12.33 sec

Horizontal distance traveled = 252×12.33cos30°

=2690.87 m

therefore the angle θ is

⇒ θ= 40.52°

y=mx+b

step-by-step explanation:

m is slope, b is y-intercept

i am thinking c. sorry if im wrong .

step-by-step explanation: