An airplane is fl ying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a fl are is released from the plane. The fl are hits the target on the ground. What is the angle θ?

Answer :

Answer:

[tex]\theta=41.52^{\circ}[/tex]

Explanation:

Given that,

Velocity of the  airplane, v = 240 m/s

Angle with horizontal, [tex]\theta=30^{\circ}[/tex]

The altitude of the plane is 2.4 km, d = 2400 m

Vertical speed of the airplane, [tex]v_y=v\ sin\theta=240\ sin(30)=120\ m/s[/tex]

Horizontal speed of the airplane, [tex]v_x=v\ cos\theta=240\ sin(30)=207.84\ m/s[/tex]

So, the equation of the projectile for the flare is given by :

[tex]4.9t^2+120t-2400=0[/tex]

On solving the above equation, we get the value of t as:

t = 13.04 seconds

Horizontal distance travelled,

[tex]d=v_x\times t[/tex]

[tex]d=207.84\times 13.04[/tex]

d = 2710.23 m

Let [tex]\theta[/tex] is the angle with which it hits the target. So,

[tex]tan\theta=\dfrac{2400}{2710.23}[/tex]

[tex]\theta=41.52^{\circ}[/tex]

Hence, this is the required solution.

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