Arocket is fired vertically from the ground. it moves upwards with a constant acceleration 10m/s square for 30 seconds after which the fuel is finished. after what time from the instant of the firing of the rockets will attain maximum height? ,find the maximum height it rraches above the ground and total time the rockrt will stay above the ground.

9090 m, 104 s

Explanation:

After the acceleration phase, the rocket reaches a height of:

x = x₀ + v₀ t + ½ at²

x = (0 m) + (0 m/s) (30 s) + ½ (10 m/s²) (30 s)²

x = 4500 m

And it reaches a velocity of:

v = at + v₀

v = (10 m/s²) (30 s) + (0 m/s)

v = 300 m/s

After the fuel runs out, the rocket goes into free fall.  The maximum height reached is:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (300 m/s)² + 2(-9.8 m/s²)(x - 4500 m)

x ≈ 9090 m

The time to reach maximum height during free fall:

v = at + v₀

0 m/s = (-9.8 m/s²) t + (300 m/s)

t ≈ 30.6 s

And the time to land from the maximum height:

x = x₀ + v₀ t + ½ at²

0 m = (9090 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t ≈ 43.1 s

So the total time is:

t = 30 s + 30.6 s + 43.1 s

t ≈ 104 seconds

be subject to a force andchange its motion

explanation:

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