Question

An airplane is flying with a velocity 290 m/s at an angle of 32 degrees to the horizon as shown in the image. When the altitude of the plane is 2360 meters above the ground a flare is released from the plane. The flare hits the ground at the point indicated in the image. What is the angle phi measured in degrees?

Answer 1

Explanation:

Given:

$V=290( \tfrac{m}{s} )$

$\theta = 30^{o}$

$d=2360m$

Vertical speed of the airplane $v_{y}=vsin\theta=290sin(30^{o})=145( \tfrac{m}{s})$

Horizontal speed of the airplane, $v_{x}=vcos\theta=290(30^{o})=251.147( \tfrac{m}{s} )$

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

$4.9t^{2}+145t-2360=0$

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

$t=11.67 seconds$

Horizontal distance travelled:

$d=v_{x} \cdot t$

$d=251.147 *11.67$

$d=2930.88m$

Let $\theta$ is the angle with which it hits the target. So:

$tan\theta= \dfrac{2360}{2930.88}$

$\theta=38.84^{o}$