A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 10.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 15.0 m/s2. In Stage 2, the acceleration is 10.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.
1) Calculate the maximum altitude. (Express your answer to two significant figures.)
2) Calculate time required to return to the surface (i. e. the total time of flight). (Express your answer to three significant figures.)

3.7 km

62.7175453187 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

The maximum altitude of the rocket is 750+875+2038.73598369 = 3663.73598369 m = 3.7 km

Time taken to reach the ground from max height 27.3301854818 seconds

Time to reach max height during free fall after stage 2 ends is 20.3873598369 s

Total time of flight is 10+5+20.3873598369+27.3301854818 = 62.7175453187 seconds

explanation: magnetic pole, region at each end of a magnet where the external magnetic field is strongest. the south-seeking pole, or any pole similar to it, is called a south magnetic pole. unlike poles of different magnets attract each other; like poles repel each other.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s