Question

Initially at rest a single-stage rocket is launched vertically from the ground. The rocket’s thrust overcomes gravity and provides the rocket a constant upward acceleration a. The fuel is exhausted 10 seconds after launch and then the motion of the rocket is free fall only due to gravity. If the total flight time is 30 s when the rocket strikes the ground, determine (a) the initial acceleration a, (b) the rocket’s impact speed as it hits the ground, (c) the height h from the ground the rockets reaches

Answer 1

a) The initial acceleration is 7.84 m/s²

b) The impact speed is 117.6 m/s

c) The height is 705.6 m

Step-by-step explanation:

a) The speed from A to B is:

v = u + at

Where

u = initial speed = 0

t = 10 s

Replacing:

v = 10t (eq. 1)

The vertical distance between A to B is:

`h=(1)/(2) at^(2) +ut=(1)/(2) a*(10)^(2)+0=50a` (eq. 2)

From B to C, the time it take is equal to 20 s, then:

`h=vt+(1)/(2) at^(2)`

Replacing eq. 1 and 2:

`-50a=(10a*20)-(1)/(2) *g*20^(2) \\-250a=-(1)/(2) *9.8*20^(2) \\a=7.84m/s^(2)`

b) The impact speed is equal:

`v_(i) ^(2) =v^(2) +2gs`

Where

s = h = -50a

`v^(2) _(i) =(10a)^(2) +2*(-9.8)*(-50a)\\v=\sqrt{100a^(2)+980a } \\v=\sqrt{(100*7.84^(2))+(980*7.84) } =117.6m/s`

c) The height is:

`v_(i) ^(2) =v^(2) +2gs\\0=(10a)^(2) -2gs\\(10a)^(2) =2gs\\s=((10a)^(2) )/(2g) \\s=((10*7.84)^(2) )/(2*9.8) =313.6m`

htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m

Answer 2

Step-by-step explanation:

Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds

s = 1/2at²

= .5 x a x 100

=50a

Velocity after 10 s

u = at = 10a

Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.

s = ut + 1/2 gt²

50a = -10ax20 + .5 x 10x400

250a = 2000

a = 8 m s⁻² .

Velocity after 10 s

= at = 80 m/s

further height reached with this speed under free fall

h = v² / 2g

= 80 x 80 / 2 x 10

= 320 m

height achieved under acceleration

= 50a

= 50 x 8 = 400m

Total height

= 320 + 400= 720 m

velocity after falling from 720 m

v² = 2gh = 2 x10 x 720

v = 120 m/s