- #1

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Thank You for you help

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- Thread starter bard
- Start date

- #1

- 65

- 0

Thank You for you help

- #2

- 43

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i take it you mean that the acceleration is constant until the fuel is burnt out.

assuming accleration is constant, we know that up to the point of fuel burn out:

initial velocity

acceleration

displacement

start by answering part (b):

since acceleration is constant we can apply the equation:

d = ut + 1/2at^2

where t = time taken to reach 1200m. substitute the above values in to give an equation in t:

1200 = 0t + 1/2(3.2)t^2

1200 = 1.6t^2

rearange to get t:

t = √1200/1.6)

t = √750

t ≈ 27.386s

now do part (a):

a is constant so we can use:

v = u + at

where v is the velocity at 1200m. substitute in above values:

v = 0 + 3.2√750

v ≈ 87.636m/s

part (c)

consider the flight of the rocket after it runs out of fuel. we know:

acceleration a ≈ -9.8m/s/s

initial velocity

velocity at maximum altitude

constant acceleration (assuming no air resistance) so use:

v^2 = u^2 + 2ad

where d is the displacement of the rocket after it runs out of fuel. substitute in values and solve for d:

0^2 = (3.2√750)^2 + 2(-9.8)d

d = (0 - (3.2√750)^2)/2(-9.8)

d = (-(3.2√750)^2)/-19.6

d = (-10.24(750))/-19.6

d ≈ 391.837m

add this displacement to the altitude of the rocket when its fuel ran out:

maximum altitude = 1200 + 391.837 = 1591.837m

part (d)

consider the flight of the rocket after it runs out of fuel. we know:

a ≈ -9.8m/s/s

u = 3.2√750m/s

v = 0m/s

use equation:

v = u + at

0 = 3.2√750 + (-9.8)t

t = (-3.2√750)/-9.8

t ≈ 8.942s

add this to the time taken to run out of fuel:

time taken to reach max altitude = 27.386 + 8.492 = 36.382s

part (e)

u = 0m/s

a ≈ 9.8m/s/s

d = 1591.837m

v = velocity at 0 altitude.

v^2 = u^2 + 2ad

v = √(u^2 + 2ad)

v = √31200.005

v ≈ 176.635m/s

total time in the air = 36.382 + 176.635/9.8 ≈ 54.406s

all of these calculations assume constant acceleration and zero air resistance.

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