consider the motion of rocket until it runs out of fuel
v₀ = initial velocity = 0 m/s
v = final velocity when it runs out of fuel = ?
t = time after which fuel is finished = 3.98 sec
a = acceleration = 29.4 m/s²
Y₀ = height gained when the fuel is finished = ?
using the kinematics equation
v = v₀ + a t
v = 0 + (29.4) (3.98)
v = 117.01 m/s
using the equation
v² = v²₀ + 2 a Y₀
(117.01)² = 0² + 2 (29.4) Y₀
Y₀ = 232.85 m
consider the motion of rocket after fuel is finished till it reach the maximum height.
Y₀ = initial position = 232.85 m
Y = final position at maximum height
v₀ = initial velocity just after the fuel is finished = 117.01 m/s
v = final velocity after it reach the maximum height = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
using the kinematics equation
v² = v²₀ + 2 a (Y - Y₀)
inserting the values
0² = (117.01)² + 2 (- 9.8) (Y - 232.85)
Y = 931.4 m
The total distance traveled by the rocket above the ground is 931.4 m.
The given parameters;
acceleration of the rocket, a = 29.4 m/s²time of motion of the rock, t = 3.98 sThe distance traveled by the rocket during the 3.98 s is calculated as follows;
[tex]h_1 = v_0t + \frac{1}{2} at^2\\\\h_1 = 0 + \frac{1}{2} (29.4)(3.98)^2\\\\h_1 = 232.85 \ m[/tex]
The final velocity of the rocket after 3.98 s is calculated as follows;
[tex]v_i= v_0 + at\\\\v_i= 0 + (29.4 \times 3.98)\\\\v_i = 117.01 \ m/s[/tex]
"when the rocket runs out of fuel, it moves at a constant speed and the acceleration is zero. The rocket will be moving against gravity."
The distance traveled by the rocket when it runs out of fuel is calculated as follows;
[tex]v_f^2 = v_i^2 - 2gh_2[/tex]
where;
[tex]v_f[/tex] is the final velocity of the rocket at maximum height = 0[tex]0 = (117.01)^2 -2(9.8)h_2 \\\\2(9.8)h_2 = (117.01)^2\\\\h_2 = \frac{ (117.01)^2}{2(9.8)} \\\\h_2 = 698.54 \ m[/tex]
Total distance traveled by the rocket above the ground;
H = h₁ + h₂
H = 232.85 m + 698.54 m
H = 931.4 m
Thus, the total distance traveled by the rocket above the ground is 931.4 m.
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