So when dealing with this type of questions we separate the motion into horizontal and vertical components and treat them independently. When dealing with the horizontal component for this projectile motion there are 2 ways:
Also considering the horizontal velocity of the plane for the motion :
x = 50 + 10 cos(25)Only considering horizontal component of the initial velocity of the particle.
x = 10 cos(25)
I think 1 is right but i dont have a good reason why. So which of them is correct and why is the other one wrong?
Imagine you are running with a ball in your hand with some velocity $u$ relative to ground. Relative to your reference frame, the ball is stationary. But, if someone on rest w.r.t. the ground observes the ball, he will see that the ball is also moving with a velocity of $u$ relative to the ground, because you are moving with that velocity and the ball is in your hand, thus moving along you.
This case is similar to the above case. The aircraft already has a velocity of $50\,\text{ ms}^{-1}$ w.r.t. ground. Thus, when the sensor is launched, the velocity of the aircraft adds with the sensor's launch velocity.
Velocity of aircraft w.r.t. ground, $\mathbf v_{a|g}$ = $50\mathbf i $.
Velocity w.r.t aircraft with which the sensor is launched, $\mathbf v_{s|a} $ =$10\cos(25^\circ)\mathbf i-10\sin(25^\circ)\mathbf j$
Thus, velocity of sensor w.r.t. ground, $\mathbf v_{s|g} = \mathbf v_{s|a}+\mathbf v_{a|g}$ $ = (50+10\cos(25^\circ))\mathbf i-10\sin(25^\circ)\mathbf j $
You can clearly see that the horizontal component of the velocity of the sensor w.r.t. ground is $50+10\cos(25^\circ)$.