The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s.
KE = PE
1/2 m • v^2 = mgh
1/2 • v^2 = gh
V^2/2 = gh
2 • (v^2/2) = 2 • (gh)
V^2 = 2gh
V = sq root (2gh).
Answer: 
Explanation:
Given
final velocity at takeoff 
Acceleration of the plane can be 
Initial velocity is zero for the plane i.e. 
Using the equation of motion
![\Rightarrow v^2-u^2=2as\quad [\text{s=displacement}]\\\text{Insert the values}\\\Rightarrow (28.1)^2-0=2\times 2\times s\\\\\Rightarrow s=\dfrac{789.61}{4}\\\\\Rightarrow s=197.40\ m](https://tex.z-dn.net/?f=%5CRightarrow%20v%5E2-u%5E2%3D2as%5Cquad%20%5B%5Ctext%7Bs%3Ddisplacement%7D%5D%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5CRightarrow%20%2828.1%29%5E2-0%3D2%5Ctimes%202%5Ctimes%20s%5C%5C%5C%5C%5CRightarrow%20s%3D%5Cdfrac%7B789.61%7D%7B4%7D%5C%5C%5C%5C%5CRightarrow%20s%3D197.40%5C%20m)
Thu,s the minimum length must be 
Remember the definition of work done.
Work done is force(F) times displacement(x)
∴ W = F.Δx
According to Newton's 2nd law of motion,
F = ma
∴ W = ma.Δx ---- (i)
Using the kinematical equation v²-u² = 2ax,
aΔx = (v²-u²)/2
Plug this value in (i),
∴W = m[

]
∴W =

Which is nothing but change in kinetic energy.
That is how kinetic energy is derived
Answer:
Explanation:
Given that,
Electric Field is
E = 650 N/C
The potential at x1 = 3 is
V1 = 1700V
What is the potential at x2 = 1
V2 =?
Electric potential is given as
V = Ed
Then,
E = V/d
Therefore, the electric field is gradient of the potential and the position.
So,
E = —∆V / ∆x
E = — (V2 — V1) / (x2 — x1)
E = —(V2—1700) / (1—3)
650 = — ( V2—1700) / -2
650 × -2 = —V2 + 1700
-1300 = -V2 + 1700
V2 = 1700+1300
V2 = 3000 V