Arocket is fired vertically and ascends with a constant vertical acceleration of 20 m/s2 for 1 min. its fuel is then all used and it continues as a free fall particle. (a) calculate the maximum altitude reached. (b) calculate the total time elapsed from takeoff until the rocket strikes the earth.

a)

b)

Explanation:

Given:

Equations of motion will be used to solve get maximum altitude and total time taken

Distance travelled by the rocket during acceleration

Let v be the velocity of the rocket at the end of 60 s which is given by

a) Now the rocket will be under the gravity after its fuel ends, at the highest point of trajectory its final velocity will be zero

During this time the rocket will rise to the maximum height under gravity

Hence the maximum height of the rocket is calculated.

b) When the rocket falls back to the ground its displacement will be zero with respect to the ground

The total time

Max height= 36000 metres

Total Time of flight = 120 sec

Explanation:

It's acceleration is 20 m/s².

Time for vertical uplifting is one minute= 60 sec

V= U - at

At Max height, final velocity = zero

0 = U - at

U = at

U= 20*60

U= 1200 m/s

Formula for Max altitude

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Max height= 36000 metres

Time of flight= 2Usintita/g

= 2*1200/20

= 2400/20

= 120 sec

2 min

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6