The correct answer to the question is : 7 m/s.
EXPLANATION:
As per the question, the apple is at a height h = 2.5 m.
The initial velocity of the apple u = 0
We are asked to calculate the final velocity [ v ] of the apple before touching the ground.
As the apple is falling under gravity, then the acceleration produced on the apple is nothing else than acceleration due to gravity.
Hence, acceleration a = g = [tex]9.8\ m/s^2[/tex]
From equation of kinematics , we know that -
[tex]v^2-u^2\ =\ 2as[/tex]
⇒ [tex]v^2=\ u^2+2gh[/tex]
⇒ [tex]v^2=\ 0^2+2\times 9.8\times 2.5[/tex]
⇒ [tex]v^2=\ 49[/tex]
⇒ [tex]v=\ \sqrt{49}\ m/s[/tex]
⇒ [tex]v=\ 7\ m/s[/tex] [ans]
Hence, the final velocity is 7 m/s.
