How High Can This Rocket Go After The Engines Shut Off?

when rocket accelerates, its high can be calculated with the following formula $y_{1}=\frac {1}{2} at^{2}$ $y_{1}=\frac {1}{2} \times 8\times 20^{2}$ $y_{1}=1600 m$

velocity when the rocket motor has shuts off can be calculated with the following formula $v_{0}=at$ $v_{0}=8\times 20$ $v_{0}=160 m/s$

so we can calculate the height when it continues to coast upwards $v^{2}-v_{0}^{2}=-2gy_{2}$ $0-160^{2}=-2\times 9.8\times y_{2}$ $\frac {160^2}{2\times 9.8}=y_{2}$ $1306.122 m\approx y_{2}$

$H=y_{1}+y_{2}$ $H\approx 1600 m+1306.122 m$ $H\approx \boxed{2906.122 m}$

We use here all the 3 motion equations $v^{2}=u^{2}+2as , v=u+at , s=ut+\frac{1}{2}at^{2}$

Here, while the motor is running, the rocket flies with acc. $(a)=8$ m/ $s^{2}$ for $t=20$ sec. and had initial velocity $(u)=0$ .

Here, $v=u+at \implies v=0+8\times 20 \implies v=160$ m/s. This is the initial velocity for the journey after motor shuts off.

Distance travelled while motor was running $=ut+\frac{1}{2}at^{2} = 0\times 20 +\frac{1}{2}\times 8\times (20)^{2} = 0+\frac{1}{2}\times 8\times 400 = \boxed{1600}$ m

Now, when motor shuts off, the acc. acting on the rocket is $(-9.8)$ m/ $s^{2}$ and initial velocity $=160$ m/s

Now, $v^{2}=u^{2}+2as \implies 0=160^{2}-2\times 9.8\times s \implies s=\frac{160\times 160}{19.6} \implies s=\boxed{1306.12}$

Now, total distance travelled $=1600+1306.12 = \boxed{2906.12}$

We must consider two steps in our solution:

1) Acceleration

During the first phase, the rocket is accelerating straight up from rest with an acceleration of $\frac {8m}{s^2}$ , during 20 seconds. Since it has no initial velocity, and a constant acceleration, we can use the following formula:

$\Delta s_1 = \frac{a \times t^2}{2}$

$\Delta s_1 = \frac{8 \times 20^2}{2}$

$\Delta s_1 = 1600$ $m$

2) Deceleration

During the second phase, the rocket no longer has an upward acceleration, while still in an ascending trajectory due to its upward speed. Let's calculate the initial speed of the second phase:

$\Delta v = a \times t$

$\Delta v = 8 \times 20$

$\Delta v = 160 \frac{m}{s}$

In this case, since there is an initial speed, we cannot use the same formula (previous step), so we must use one that considers speed:

$\Delta s_2 = v \times t + \frac{a \times t^2}{2}$

$\Delta s_2 = 160 \times t + \frac{(-9.8) \times t^2}{2} (1)$

In order to calculate the second displacement, the duration of the second phase must be known. It can be found by calculating the time for the rocket to reach zero velocity:

$\Delta v = a \times t$

$v_f - v_i = (-9.8) \times t$

$0 - 160 = (-9.8) \times t$

$t = \frac {160}{9.8} s$

Knowing the amount of time, the second displacement can be found by returning to equation (1):

$\Delta s_2 = 160 \times t + \frac{(-9.8) \times t^2}{2} (1)$

$\Delta s_2 = 160 \times \frac{160}{9.8} + \frac{(-9.8) \times \frac{160}{9.8}^2}{2}$

$\Delta s_2 = \frac{12800}{9.8}$ = $1306.12$ $m$

3) Conclusion

$h = \Delta s_1 + \Delta s_2$

$h= 2906.12$ $m$

First find the distance travelled in the first twenty seconds using the equation of motion $S=ut + \frac{1} {2} a t^2=0.5 \times 8 \times 20^2=1600~m$ since the initial velocity $u$ is zero and we know $a$ . We can then find the final velocity for the first part of the motion via $V=u+at=0+8 \times 20 =160~m/s$ . For the second part of the motion this velocity is now considered as the initial velocity and the acceleration is $-9.8~m/s^2$ . The final velocity is zero because the rocket will stop at the top of its motion. Now find the distance travelled in the second part of motion via $V^2=u^2+2as, 0^2=160^2-2 \times 9.8 \times s$ which yields s=1306 m. Therefore the total distance travelled will be s=1600+1306=2906 m.

Considering the first 20 seconds, the initial velocity, u=0ms^-1, acceleration, a=8ms^-2 (given).

Using the kinematic equation, displacement, s= u \times t+ \frac{1}{2} \times a \times t^2, we can find the displacement in the first 20 seconds which equals 1600m. The final velocity, v can be found using the equation v=u+a \times t= 160ms^-1.

Considering the following part of the flight, the initial speed is 160ms^-1, obtained from the final velocity in the first 20 seconds. Acceleration is 9.8ms^-2 downwards and the final velocity is now 0. Using the equation v^2=u^2+2\dot a \dot s, we can obtain the displacement to be approximately equal to 1306.

Summing the two respective displacements, we will obtain the total vertical displacement 2.91E3 to 3 significant figures.

First Vf=Vi + at Vf=0 + 8(20) = 160m/sec

To know distance during the first 20 secs: Vf^2 = Vi^2 + 2as 160^2 = zero + 2(8)(s) S1=1600 m

Second ( motor shuts off ) Vf^2 = Vi^2 + 2as zero = 160^2 - 2(9.8)(S) S2= 1306 m

Total distance = S1 + S2 = 2906m

for the first 20 seconds the rocket accelerates upwards with 8 m/s^2. let it attain a velocity Vo after t=20 sec

Vo = Uo+a*t where Vo is final velocity ; Uo is the initial velocity ; a is the acceleration of body ; t is the time for which the body moves under the acceleration a

Vo = 0 + 8*20 = 160m/s

in this interval it will move by a distance (S)

S=Uo t + .5 a t^2 = 0 20 + .5 8 20*20 = 1600 m

after 20 seconds it will move freely under gravity.

distance (S1) it will move before coming to rest

Vo^2-Uo^2 = 2 a (S1)

here Vo = 0 as it comes to rest at topmost point Uo = 160 m/s a = -g = -9.8m/s^2 (negative as it is retarding the motion)

0-160^2= 2 (-9.8) (S1) S1 = 1306.12 m

total distance travelled = S + S1 = 1600m + 1306.12m = 2906.12 m

First and for most, the rocket starts from an initial speed , $u$ , of 0 $m/s$

Using the kinematics formula

$s = ut + \frac{1}{2}at^{2}$ , given $a$ = $8$ $m/s^{2}$ , $t$ = $20$ $s$

$s_{1} = (0)(20) + \frac{1}{2}(8)(20)^{2}$

$s_{1} = 1600$ $m$

denoting the distance travelled by the rocket for the first $20$ $s$

After the acceleration stops, the rocket thus experiences a deceleration of $9.8$ $m/s^{2}$ due to the effect of gravity.

Thus, given $a$ = - $9.8$ $m/s^{2}$

Firstly, we need to work out the final speed of the rocket after 20 seconds of acceleration, $v_{1}$ , that is equals to the initial speed of the decelerating of the rocket due to gravity , $u_{2}$

Thus,

$v$ = $u$ + $at$

$v_{1}$ = $0$ + $(8)(20)$

$v_{1}$ = $160$ $m/s$ = $u_{2}$

Thus, with $u_{2}$ = $160$ $m/s$ , $a$ = - $9.8$ $m/s^{2}$

Using the kinematics formula

$v^{2}$ = $u^{2}$ + $2as$

$0$ - $160^{2}$ = $2(-9.8)(s_{2})$

$s_{2}$ = 1306 m (approx.)

Hence, total height travelled :

$H$ = $s_{1}$ + $s_{2}$

$H$ = $1600$ + $1306$

$H$ = $2906$ $m$

Using Kinematics Formulas:

When rocket motor is running

s=ut+ 1/2 a $t^{2}$

Where:

• u (Initial velocity) = 0m $s^{-1}$
• t (Time) = 20s
• a (Acceleration) = 8m $s^{-2}$

Substitute the values in and find s which is 1600m

After the rocket stops, the rocket travels at 160m $s^{-1}$

$v^{2}$ = $u^{2}$ + 2as

Where:

• u (Initial velocity) = 160m $s^{-1}$
• v (Final velocity) = 0m $s^{-1}$
• a ( Gravitational Acceleration) = -9.8m $s^{-2}$

Substitute the values again and find s which is 1306 (1306.12244)

Total distant traveled by rocket = 1306 + 1600 = 2906

since rocket starts from rest - so , u = 0 by H = ut + 1/2 at ^ 2 = 0 * t + 1/2 * 8 * ( 20 ) ^ 2 = 0 + 1600 = 1600 m and final velocity at this point = u + at = 0 + 8 * 20 = 160 m / s

when the motor shuts off - the acceleration due to gravity will act and the final velocity for 1 st case will be initial velocity for 2nd case and final velocity at the top will be 0 . So by . v ^ 2 - u ^ 2 = 2 a s
=> 0 ^ 2 - ( 160 ) ^ 2 = 2 * - 9.8 * h => 0 - 25600 = - 19.6 h

By solving , we get h= 1306 .122 m

Adding the 2 hts . in both the cases = 2906 .122 m

distance of the rocket at the end of twenty seconds=0.5 a t t=1600.. speed at the end of 1600m=v v =2 a s=160m/s after the engine shuts off acceleration of the rocket =acceleration due to gravity initial speed of the rocket=160m/s final speed of the rocket=0 h=? (v v)/(2*a)=s by putting the values you wull get the value of h as 1306. add initial height 1600

you will get the required height

Case 1: Motion of the rocket for first 20 sec by using s=ut+1/2at^2 we get - h=0+1/2 8 20^2 or, h=1600 m let, velocity after first 20 sec be v then by using v=u+at we get - v=0+8 20 or, v=160 m/s Case 2: Motion of the rocket after first 20 sec by using v^2=U^2+2as we get - 0=160^2-2 9.8*s or, s=1306.12 m Therefore, H=h+s=2906.12 m

$s_1 = V_o \cdot t+ \frac {1}{2} \cdot a \cdot t^2 =>$ $s_1 = 0 \cdot 20 + \frac {1}{2} \cdot 8 \cdot 20^2$

$s_1 = \frac {1}{2} \cdot 3200 =>$ $s_1 = 1600$ Now let's find the speed when it lost it's acceleration :

$V_t = V_o + a \cdot t => V_t = o + 160 => V_t = 160$

$V_t^2 = V_o^2 + 2 \cdot a \cdot s_2 => 0 = 160^2 + 2 \cdot -g \cdot s_2$

$0 = 25600 - 18.6 \cdot s_2 => 18.6 \cdot s_2 = 25600 => s_2 = \frac {25600}{18.6}$

$s_2 = 1306 => s = s_1 + s_2 => s = 1600 + 1306 = 2906$

We can find the speed of the rocket when the engines shutdown, from v=u+at (u=0) and the height it is at from, s=ut+1/2at^2 (u=0). By substituting the values given in the question, we get v=160m/s and s=1600m------(1)

From there, we can get the height reached by the rocket from the point where the engines shut off from, v^2-u^2=2as (v=0, since, if it had velocity at the top most point it would go further high). We get from there s=1306.12m.-------(2)

Adding both heights from Eq.s (1) and (2) we have the max. height reached i.e., 1600m+1306.12m=2906.12m

The question asks for how high the rocket goes in meters after accelerating at 8m/s^{2} for 20 seconds. The answer will be acquired by combining the distance the rocket travels while undergoing acceleration, as well as the distance traveled while the rocket coasts upward.

In order to find the distance traveled while the rocket is accelerating, we can use the formula S=U+/frac{1}{2}AT^{2} , whereby T= Time, U= Initial velocity, A= Acceleration, S= Distance traveled,

S=0+/frac{1}{2}8(20)^{2}

S=1600m

Now, to find the distance traveled when the rocket stops accelerating, we can use the equation V=U+AT , whereby V= final velocity

However, we don't have the new initial velocity when the rocket stops accelerating. So, we will have to substitute the numbers from when the rocket was still accelerating, in order to find the new initial velocity for when the rocket stops accelerating.

V=U+AT

V=0+(8)20

V=160m/s

Now that we have the initial velocity for when the rocket stops accelerating, we can find the distance traveled while the motor to the rocket is shut off.

Using the new equation, V^{2}=U^{2}+2AS

0=160^{2}+2(-9.8)S

S=1306

Combining the distances traveled while the rocket was accelerating and stopped accelerating, 1306+1600=2906m

For this, we have to use a set of kinematic equations. $s = ut+\frac{1}{2}at^{2}$ $v=u+at$ $v^2=u^2+2as$ We try to split the calculations into 2 stages. The first stage is the rocket in the 20s burst, the second stage is the rocket after the 20s burst.

First, we need to find the distance travelled. We use the first 2 eqns for this.

$s_{1} = (0)(20)+\frac{1}{2}(8)(20)^{2} = 1600m$

We realise that the final speed of the rocket after 1st stage is the same as the starting speed of the rocket of the 2nd stage. We use the 2nd eqn to find the starting speed of the rocket in the 2nd stage, $u_{2}$

$v_{1} = u_{2} = 0+8(20) = 160m/s$

Knowing this speed, we can now find the distance the rocket travelled in the 2nd stage using eqn 3:

$0^{2} = 160^{2} + 2(9.8)s$

$s_{2} = \frac{160^{2} }{ 2(9.8)} = 1306.122$

The final answer is thus $s_{1}+s_{2} = 2906m$

When accelerating the rocket goes up a distance $h_{1} = \frac{a t_{1}^{2}}{2}$ , where $t_{1}$ is the time that the rocket spends accelerating up and $a$ is its acceleration during the proccess. When it is not accelerating (up) anymore, it continue going up, but accelerating down (because of gravity). If we consider the gravity acceleration a constant during all the time, Torricelli's equation yield $0 = (a t_{1})^{2} - 2 g h_{2}$ where $h_{2}$ is the distance it goes up before stopping at the highest point of its trajectory. Then, obviously $H = h_{1} + h_{2} = \frac{a t_{1}^{2}}{2} (1 + a/g) \approx 2906.000 m$

v = Final Velocity, u = Initial Velocity, s = Distance, t = Time, a = acceleration

first, u = 0 (since rocket starts from rest), a = 8, t = 20 & v = ?, s = ?

since (v - u)/t = a, then (v - 0)/20 = 8. Thus v = 160. Also, since s = ut + (1/2)(a(t^2)), s = 1600

Here 160 is the final velocity and rocket has covered 1600m during Engine ON. In the next phase, 160 is treated as initial velocity (i.e. u)

Thus, u = 160, v = 0, s = ?, a = -9.8

(v^2) - (u^2) = 2as

Therefore, -(160^2) / (-9.8 x 2) = s = 1306.1224

Hence, total distance = 1306 + 1600 = 2906.

Key to solve this is the equations of Kinematics and preferably a biological equipment called brain.

From the ground to where the engines stop:

$\Delta x = v_0t+\dfrac{1}{2}at^2 = \dfrac{1}{2}(8)(20)^2 = 1 600$ .

The speed at the top is thus $v^2 = v_0^2 + 2a\Delta x \implies v = \sqrt{2\cdot 8 \cdot 1600 } = 160$ .

From there, to the highest point: $v^2 = v_0^2 + 2a\Delta x \implies 0 = 160^2 + 2(-9.8)(\Delta x) \implies \Delta x = 1306$ .

Adding this to the first height we get: $1600+1306 = \boxed{2906}$ .

The first part : discover how High does the rocket when ir is accelerating. So : S = So+Vo.t+a.t^2/2 =>

S = 0 + 0.20 + 8.(20.20)/2 =>

S = 1600 meters.

Second part : how High here got when the engine sutis Office,So:

V^2= Vo^2 - 2.4,9.S =>

0 = 160^2 - 9,8S => S = 1300. => Now we have to sum the two parts,So :

S is aproximately 2900 meters.

Let $t=20 \text{ s}$ .

$v(t)=v(0)+at$

$v(t)=0+(8\text{ m/s})(20\text{ s})=160\text{ m}$

$h(t)=\frac{1}{2}at^2+v(0)t+s(0)$

$h(t)=\frac{1}{2}(8\text{ m/s}^2)(20\text{ s})^2+0+0=1600\text{ m}$

Let $a$ be the time at which the rocket reaches its maximum height.

$v(a)^2=v(t)^2+2a(h(a)-h(t))$

$0=(160 \text{ m/s})^2-2(9.8 \text{ m/s}^2)(h(a)-h(t))$

$h(a)-h(t)=1306 \text{ m}$

$H=h(a)=1306 \text{ m}+1600 \text{ m} = \fbox{2906} \text { m}$