Answer:
The speed of the plane before the collision is 12.6 m/s.
Explanation:
Let us get some things clear first.
The drone is hovering stationary at an altitude of 11.3 m, which means its velocity is zero. Furthermore, after the collision the two aircraft are stuck together and move east at an altitude of 11.3 m, which means their velocity has no vertical component because if it had their altitude would have changed.
The law of conservation of momentum says that
[tex]m_1v_1+m_2v_2= m_1v_{f1}+m_2v_{f2}[/tex].
where [tex]m_1 =2.4kg[/tex] and [tex]m_2= 1.2kg[/tex].
Since [tex]v_2 = 0[/tex] (the drone is hovering stationary), and [tex]v_{f1}= v_{2f}= 8.3m/s[/tex] (the aircraft are stuck together), we have
[tex]2.4v_1 = (8.3m/s)(2.4kg+1.2kg)[/tex]
[tex]v_1= \dfrac{(8.3m/s)(2.4kg+1.2kg)}{2.4}[/tex]
[tex]\boxed{v_1 = 12.6m/s}[/tex]
Thus, the speed of the plane prior to the collision is 12.6 m/s.
