# A rocket flies directly upward with an initial velocity of 24.0 m/s. Assuming there is no air friction, what is the maximum height

###### Question:

## Answers

151 ft

48 ft/s

Step-by-step explanation:

h(t)=-16t^2+v₀t+h₀

v₀ = 96 and

h₀ = 7

h(t)=-16t^2+96t+7

The maximum height is at the vertex

The x values of the vertex is at

x = -b/2a = -96/ (2*-16) = -96/-32 =3

Substitute this into the function to find the maximum height

h(3) = -16 ( 3)^2 +96*3+7

-16*9 +288+7

151

The maximum height is 151 ft

Average rate of change from 0 to 3

h(3) - h(0)

3-0

151 -7

3

144/3

48ft/s

a) 112.5 m

b) 15.81s

Explanation:

a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2

[tex]v^2 = 2as[/tex]

[tex]v^2 = 2*2.25*500 = 2250[/tex]

[tex]v = \sqrt{2250} = 47.4 m/s[/tex]

After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure

[tex]0^2 - v^2 = 2gh[/tex]

[tex]-2250 = 2(-10)h[/tex]

[tex]h = 2250/20 = 112.5 m[/tex]

So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m

b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:

[tex]s + vt + gt^2/2 = 0[/tex]

[tex]500 + 47.4t - 10t^2/2 = 0[/tex]

[tex]5t^2 - 47.4t - 500 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{47.4\pm \sqrt{(-47.4)^2 - 4*(5)*(-500)}}{2*(5)}[/tex]

[tex]t= \frac{47.4\pm110.67}{10}[/tex]

t = 15.81 or t = -6.33

Since t can only be positive we will pick t = 15.81s

a) 30.3 m/s b) 46.8 m

Explanation:

a)

Assuming no friction present, the work-energy theorem says that the work done on the rock by the only force acting on it (gravity force, aiming downward), is equal to the change of kinetic energy of the rock, at any point.When it's at 15.0 m high, the work done by gravity on the rock can be written as follows:[tex]W = m*g* h* cos 180\º = -20 N* 15.0 m = -300 J (1)[/tex]

The change in kinetic energy of the rock can be expressed in this way:[tex]\Delta K = K_{f} - K_{0} = \frac{1}{2} * m* (v_{f}^{2} -v_{0} ^{2} ) (2)[/tex]

where m= 2.04 kg, vf = 25.0 m/s.As (1) and (2) are equal each other, we can solve for v₀, as follows:[tex]= -300 J = 0.5*2.04 kg *(( 25.0 m/s)^{2} -v_{0}^{2} )\\ \\ v_{0}^{2} = \frac{300*2 J}{2.04kg} + 625 (m/s)2 = 894. 1 (m/s)2 \\ \\ v_{0} = \sqrt{894.1 (m/s)2} = 30.3 m/s[/tex]

b)

When the rock arrives to its highest point, it will momentarily come to a stop, before changing direction to start falling.So, the final kinetic energy, will be zero.Applying the work --energy theorem again, we can write the following equation:[tex]\Delta K = K_{f} - K_{0} =( 0- \frac{1}{2} *m*v_{0}^{2}) = -m*g*h_{max}[/tex]

Replacing by v₀ = 30.3 m/s, rearranging and simplifying common terms, we can solve for hmax, as follows:[tex]h_{max} = \frac{v_{0} ^{2} }{2*g} =\frac{(30.3 m/s)^{2}}{2*9.8 m/s2} = 46.8 m[/tex]

A) the maximum height this rocket will reach above the launch pad is 614.68 m

B) time elapsed after engine failure before the rocket comes crashing down to the launch pad is 16.025 sec.

Explanation:

Detailed explanation and calculation is shown in the image below

[tex]A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

[tex]A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

[tex]A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

Explanation:

Given

Initial velocity (u)=0

t=15.4 s

acceleration(a)=[tex]2.25 m/s^2[/tex]

(a)maximum height the rocket will reach is s

v=u+at

[tex]v=0+2.25\times 15.4=34.65m/s[/tex]

[tex]v^2-u^2=2gs[/tex]

[tex]0-34.65^2=-2\times 9.81\times s[/tex]

s=61.19 m

(b) velocity of rocket before it crashes on to launchpad

[tex]v^2-u^2=2as[/tex]

Now rocket will be under complete control of gravity

here u=0 as rocket has reach its maximum height

up to 15.4 s rocket has covered a distance of 266.805

[tex]v^2=2\times 9.81\times (61.19+266.805)[/tex]

[tex]v=\sqrt{6435.261}=80.22 m/s[/tex]

(c)Time taken will be addition of rocket reaching maximum height plus rocket reaching launchpad

[tex]t_1=\frac{34.65}{9.81}=3.53 s[/tex]

Now from maximum height to launchpad

[tex]v=u+at_2[/tex]

[tex]80.22=0+9.81\times t_2[/tex]

[tex]t_2=8.177 s[/tex]

t=3.53+8.177=11.71 s

Explanation:

initial vertical velocity = 17.5 m/s

using g=-9.81 m/s^2

apply kinematics equation

v1^2-v0^2=2gS

solve for S with v1=0, v0=+17.5

S = (v1^2-v0^2)/2g

=(0-17.5^2)/(2*(-9.81))

= 15.61 m

H = 645.42 m

v_4 = 112.53 m/s

T = 16.43 s

Explanation:

Given:-

- The mass of the rocket m = 7500 kg

- The acceleration for first part of the journey a_1 = 2.25 m/s^2

- The height reached before engine failure h_1 = 525 m

Find:-

(a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?

(c) Sketch ay−t,vy−t, and y - t graphs of the rocket’s motion from the instant of blast-off to the instant just before it strikes the launch pad.

Solution:

- Find the velocity v_2 when the engine fails. Using 3rd equation of motion.

v_2^2 = v_i^2 + 2*a_1*h_1

The rocket blasts off from rest v_i = 0.

v_2 = sqrt ( 2*a_1*h_1 )

v_2 = sqrt ( 2*2.25*525 ) = sqrt ( 2362.5 )

v_2 = 48.606 m/s

- The time taken for first part of upward journey t_1. Using first equation of motion:

v_2 = v_1 + a_1*t_1

t_1 = v_2 / a_1

t_1 = 48.606 / 2.25

t_1 = 21.603 s

- Now for second part of the upward journey, the initial velocity is v_2 and final velocity is when rocket reaches maximum height v_3 = 0. The only force is gravity; hence, a_2 = g = -9.81 m/s^2. Using 3rd equation of motion.

v_3^2 - v_2^2 = 2*a_2*h_2

h_2 = v_2^2 / 2*g

h_2 = 2362.5 / 2*9.81 = 120.415 m

- The maximum height the rocket will reach above launch pad is H:

H = h_1 + h_2

H = 525 + 120.415

H = 645.42 m

- The time taken t_2 for second part of upward journey, Using first equation of motion is:

v_3 = v_2 + a_2*t_2

t_2 = v_2 / g

t_2 = 48.606 / 9.81

t_2 = 4.955 s

- The final velocity v_4 as soon as the rocket crashes down. Using 3rd equation of motion.

v_4^2 = v_3^2 + 2*a_2*H

v_4 = sqrt ( 2*a_2*H )

v_4 = sqrt ( 2*9.81*645.42 ) = sqrt ( 12663.1404 )

v_4 = 112.53 m/s

- The time taken t_3 for downward part of journey, Using first equation of motion is:

v_4 = v_3 + a_2*t_3

t_3 = v_4 / g

t_3 = 112.53 / 9.81

t_3 = 11.47 s

- The total Time taken after engine failure to crash is T:

T = t_2 + t_3

T = 11.47 + 4.955

T = 16.43 s

[tex]A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

[tex]A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

[tex]A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.[/tex]

maximum height is 72.09 m which can be rounded to 72.1 m (Answer B)

Explanation:

In order to analyze the vertical motion and the maximum height the rocket reaches, we need to know the vertical component of the velocity at launch. That can be estimated knowing the launching angle of 70 degrees using right angle trigonometry:

[tex]v_y=40\,*\,sin(70^o)\\v_y=37.59\,\frac{m}{s}[/tex]

We recall that when the rocket reaches it maximum vertical height, its vertical component of the velocity is zero, and from then on, the trajectory of the rocket is going to go downwards accelerated by the action of gravity. We can use the following equation that relates initial and final velocities, acceleration (g), and distance traveled (final vertical position minus initial):

[tex]v_f^2-v_i^2=2\,a\,(y_f-y_i)\\0-37.59^2=2\,(-9.8)\,(y_f-y_i)\\\frac{-1413}{-19.6} =(y_f-y_i)\\72.09\,m=(y_f-y_i)\\(y_f-0)=72.09\,m\\y_f=72.09\,m[/tex]

This is therefore the maximum height at the end of the upwards motion, when the vertical component of the velocity becomes zero.

Height at failure, h1 = 525 m,

upward acceleration, a = 2.25 m/s^2,

velocity = v m/s,

SO,

v^2 = 2*a*h = 2*2.25*525 = 2362.5

Now, acceleration, g = 9.8 m/s^2,

SO,

heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters

Hence,

a)

Total height = 525+120.54 = 645.54 meters

b)

time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec

Here,

height at failure, h1 = 525 m,

upward acceleration, a = 2.25 m/s^2,

velocity = v m/s,

SO,

v^2 = 2*a*h = 2*2.25*525 = 2362.5

Now, acceleration, g = 9.8 m/s^2,

SO,

heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters

Hence,

a)

Total height = 525+120.54 = 645.54 meters

b)

time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec

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