Answer to Question #136800 in Physics for Whylist

Question

A rocket moves straight upward, starting from rest with an acceleration of 29.4 m/s2. It runs out of fuel at the end of 4s and continues to coast upward, reaching a max height before falling back to Earth. 

(a) Find the rocket’s velocity and 

position at the end of 4 s. 

(b) Find the rocket’s max height.

(c) Find it’s velocity before crashing to the ground.

Answer 1

a) We can find the rocket's velocity and position from the kinematic euations:

$$v = v_0 + at,$$ $$y = v_0t + \dfrac{1}{2}at^2,$$

here, $v_0 = 0 \ \dfrac{m}{s}$ is the initial velocity of the rocket, $a = 29.4 \ \dfrac{m}{s^2}$ is the acceleration of the rocket, $t = 4 \ s$ is the time during which the rocket runs out of fuel.

Then, we get:

$$v(t=4 \ s) = 29.4 \ \dfrac{m}{s^2} \cdot 4 \ s = 117.6 \ \dfrac{m}{s},$$ $$y(t=4 \ s) = \dfrac{1}{2} \cdot 29.4 \ \dfrac{m}{s^2} \cdot (4 \ s)^2 = 235.2 m.$$

b) Let's first find the distance traveled by the rocket from the end of 4s until it reaches max height:

$$v_f^2 = v_i^2 + 2gy_2,$$

here, $v_f = 0 \ \dfrac{m}{s}$ is the final velocity of the rocket, $v_i = 117.6 \ \dfrac{m}{s}$ is the initial velocity of the rocket, $g = -9.8 \ \dfrac{m}{s^2}$ is the acceleration due to gravity, $y_2$ is the distance traveled by the rocket from the end of 4s until it reaches max height.

Then, from this equation we can find $y_2$:

$$y_2= \dfrac{v_f^2-v_i^2}{2a}=\dfrac{(0 \ \dfrac{m}{s})^2-(117.6 \ \dfrac{m}{s})^2}{2 \cdot (-9.8 \ \dfrac{m}{s^2})} = 705.6 \ m.$$

Finally, we can find the rocket's max height:

$$h_{max} = y_1 + y_2,$$

here, $y_1 = 235.2 \ m$ is the distance traveled by the rocket at the end of 4s.

Therefore, we get:

$$h_{max}= 235.2 \ m + 705.6 \ m = 941 \ m.$$

c) We can find the rocket velocity before crashing to the ground from the kinematic equation:

$$v_f^2 = v_i^2 + 2gh_{max},$$ $$v_f = \sqrt{v_i^2 + 2gh_{max}},$$ $$v_f = \sqrt{(0 \ \dfrac{m}{s})^2 + 2 \cdot 9.8 \ \dfrac{m}{s^2} \cdot 941 \ m} = 135.8 \ \dfrac{m}{s}.$$

Answer:

a) $v(t=4 \ s) = 117.6 \ \dfrac{m}{s}, y(t=4 \ s) = 235.2 m.$

b) $h_{max}=941 \ m.$

c) $v_f =135.8 \ \dfrac{m}{s}.$