Question

a.rocket is fired vertically up from the ground with a resultant vertical acceleration of 10 m /s^2 . the fuel is finished in 1 minute and it continues to move up. find max height reach and time for which it continues its upward motion.​

Answer 1

Answer:

Maximum height attained by Rocket

= 36 km

Step by step explanations :

given that,

a.rocket is fired vertically up

from the ground with a resultant vertical acceleration of 10 m/s²

here,

initial velocity of the rocket = 0 m/s

[It was initially at rest]

acceleration of the rocket = 10 m/s²

time for which it goes with fuel

= 1 minute

= 60 seconds

now,

we have,

initial velocity(u) = 0 m/s

acceleration(g) = 10 m/s²

time taken(t) = 60 seconds

let the height reached by rocket with fuel be h

by the gravitational equation of motion,

h = ut + ½ gt²

putting the values,

h = 0(60) + ½ × 10 × 60 × 60

h = 18000 m

so,

height reached by rocket by its fuel = 18000 meter

now,

since it will have momentum of its motion so it will attain more height

so,

after reaching at the height of 18000

it's velocity = final velocity of the earlier motion

so,

v = u + gt

v = 0 + 10(60)

v = 600 m/s

so,

here,

we have,

initial velocity(u) = 600 m/s

final velocity(v) = 0 m/s

[It will stop for a moment]

gravitational acceleration(g) = -10 m/s²

also by the gravitational equation of motion,

v² = u² + 2gh

again putting the values,

0² = 600² + 2(-10)h

-20h = -360000

h = -360000/-20

h = 18000 meter

so,

height reached by the rocket after finishing fuel = 18000 meter

now,

total height = 18000 + 18000

= 36000 meter

= 36 km

Maximum height attained by Rocket

= 36 km

Answer 2

Answer:

Maximum height achieve by rocket is

36 km.

Solution:

Given that,

=> Initial velocity of the rocket (u)is

0m/s.

{Because, It was at rest initially.}

=> Acceleration of the rocket (g) is

10m /s²

=> Time taken (t) in which it goes with the fuel is:

60 seconds= 1 minute.

Now,

Let the height achieve by rocket be "h"

We know that,

Gravitational equation of motion,

$\sf\bigstar ut +\dfrac{1}{2}gt^{2} = h \bigstar$

Plug the given values in equation,

=> 0+(60)+ ½ × 10 × 60 × 60= h.

=> 18000 m.

Hence,

Hence,It implies that height achieve by rocket with its fuel is 18000 m.

So,

It have momentum of the

After reaching at the height of 18000

So, It's velocity = Final velocity if the earlier motion.

So,

=> v = u + GT

=> 0 + 10(60)

=>Final velocity = 600m/s.

Hence,

Initial velocity(u) = 600 m/s.

Final velocity (v) = 0 m/s.

Gravitational acceleration (g) = –10m/s²

We also know that,

Gravitational equation of motion,

$\sf \: \bigstar v^2 = u^2 + 2gh \bigstar$

Put the given values,

=> 0² = 600² + 2(-10)h.

=> –20h = – 360000.

=> -360000/-20 = 18000

so,

Total height = 18000² =18000+1800

=> 36000m = 36km.

Hence,

Maximum height achieve by rocket is 36km.