Explanation:
Initial speed of the rocket, u = 0
Acceleration of the rocket, 
Time taken, t = 3.39 s
Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :
Let x is the initial position of the rocket. Using third equation of kinematics as :
Let 
is the position at the maximum height. Again using equation of motion as :
Now 
and v and u will interchange
x = 524.14 meters
Hence, this is the required solution.
Answer with explanation :
The negative sign means that the potential energy decreases by the movement of the electron.
negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.
Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
