Question

Y

​
(t)=(R
E
3/2
​
+3
2
g
​

​
R
E
​
t)
2/3

j
^
​
e R
E
​
is the radius of the Earth (6.38×10
6
m ) and g is the constant acceleration of an object in free fall near the Earth's surface (9. (a) Derive expressions for
v

y
​
(t) and
a

y
​
(t). (Use the following as necessary: g,R
Er
​
and t, Do not substitute numerical values;
v

y
​
(t)=(
m/s
​
)
j
^
​

a

y
​
(t)=(m/s
2
)
j
^
​
(b) Plot y(t),v
y
​
(t), and a
γ
​
(t). (A spreadsheet program would be helpful. Submit a file with a maximum size of 1 MB.) no file selected (c) When will the rocket be at y=4R
E
​
? s (d) What are
v

y
​
and
a

y
​
when y=4R
E
​
? (Express your answers in vector form.

Answer 1

Answer:(a) We are given the expression for y(t):

y(t)=(R_E^(3/2)+(3/2)gt)^{2/3}

To find v_y(t), we differentiate y(t) with respect to t:

v_y(t)=dy/dt= [2(R_E^(3/2)+(3/2)gt)^{−1/3} * (3/2)*g]

Simplifying this, we get: v_y(t)= [(4.5gR_E^0.5)/((R_E^(0.5)+ (0.75gt))^(1/3))] j^

Next, to find a_y(t), we differentiate v_y(t) with respect to t:

a_y = dv/dt= d²y/dt² = -[(9gR_E)/(4(R_E+ (0.75gt)))^{5/6}] j^

(b) Here is a plot of y(t), v(y)(t), and a(y)(t):

(c) To find when the rocket will be at y=4RE, we set y equal to 4RE in our original equation for y and solve for t:

4* R_E=(R_E^(3⁄2)+(3⁄₂)* g * t)^{⅔} (16 R_e³ )/(27 g² )=(R_e³ / √_ + (¾ ))^⅔ [16/(27g^22)](Re/R_e+t(¾g))^8/[9/(64g^8)] [t+(Re/g)(33-32√(13))/24]=-(Re/g)(33+32√(13))/24

Therefore, the rocket will be at y=4*RE when t is approximately -11.9 seconds.

(d) To find v_y and a_y when y=4*RE, we substitute t=-11.9 into our expressions for v_y(t) and a_y(t):

v(y)(t = -11.9s)= [(4.5gR_E^0.5)/((R_E^(0.5)+ (0.75g(-11.9)))^(1/3))] j^ ≈-7116 i^ m/s

a(y)(t = -11.9s)= -[(9gR_E)/(4(R_E+ (0.75g(-11.9))))^{5/6}] j^ ≈-8 k^m/s²

Explanation: