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What is the height of the rocket above the surface of the earth at t=10.0 s ?

– A rocket initially at rest starts its upward motion from the earth’s surface. The vertical acceleration in +y upward direction in first $10.0s$ of flight is represented by $a_y=(12.8\frac{m}{s^3})t$.

– Part(a) – At what elevation will the rocket be at $10.0s$ from the surface of the earth?

– Part(b) – When the rocket is $325m$ above the earth’s surface, calculate its speed.

In this question, we have to find the height and speed of the rocket by integrating the acceleration with the limits of time.

The basic concept behind this question is the knowledge of the kinematics equation of acceleration, integration, and limits of integration.

Expert Answer

Integrate the kinematics equation as follows:

\[ v_y=\int_{0}^{t}{a_y}{dt} \]

Now putting the value of $t$ here which is $t=10$:

\[ v_y=\int_{0}^{10}{a_y}{dt}\]

Now putting the value of $a$ here which is given $a=2.8t$:

\[ v_y=\int_{0}^{10}{2.8t}{dt} \]

Now integrating the equation we get:

\[ v_y=2.8(t^ 2)(\dfrac{1}{2})+v_0 \]

Here $v_o$ is the constant which comes after the integration:

\[ v_y = 1.4 t^ 2 + v_0 \]

Here we know that $v_o=0$:

\[ v_y=1.4t^2+(0) \]

\[ v_y=1.4t^2 \]

We also know that:

\[ y=\int_{0}^{10}{v}{dt} \]

Putting $v = 1.4t^2$ in the above equation we get:

\[ y=\int_{0}^{10}{1.4t^2}{dt} \]

Taking derivative we get:

\[ y=1.4(t^3)(\dfrac{1}{3})+y_0 \]

Here we know that $y_0=0$:

\[ y=1.4[ (t^3)(\dfrac{1}{3})]_{0}^{10} + (0) \]

\[ y=\dfrac{1.4}{3}\times [ t^3 ]_{0}^{10} \]

\[ y=0.467 \times [ t^3 ]_{0}^{10} \]

Now substituting the limit of $ t$ in the above equation:

\[ y = 0.467   \times [ (10)^3 – (0)^3 ] \]

\[ y = 0.467   \times [ (10)^3  ] \]

\[ y = 0.467   \times (1000)  \]

\[ y = 467 \space m   \]

(b) Given we have $ y = 325 \space m $

we know that:

\[ y = \int { v }{ dt } \]

putting $ v = 1.4 t^ 2 $ in the above equation we get:

\[ y = \int {  1.4 t^ 2}{ dt } \]

Taking derivative we get:

\[ y = 1.4  (t^3 ) (\dfrac{1}{3} ) + y_0 \]

here we know that $ y_0 =0 $:

\[ y = 1.4 [ (t^3 ) (\dfrac{1}{3} ) ]  + (0) \]

\[ y = 1.4  [ (t^3 \dfrac{1}{3} ) ] \]

\[ y = \dfrac{1.4 }{3}   \times [ t^3 ]  \]

\[ y = 0.467   \times [ t^3 ] \]

Now substituting the value of $ y $ in the above equation, where $ y = 325 $:

\[ 325 = 0.467 \times [ t^3 ] \]

\[ 325 = 0.467   \times t^3  \]

\[ t =8.86 s \]

Putting it within the limits of the integral we have:

\[ v_y = \int_{0}^{8.86} { 2.8} { dt }\]

\[ v_y = 110 m\]

Numerical Results

(a) \[y = 467 \space m\]

(b) \[v_y = 110 m\]

Example

What is the speed of the rocket in the above question when it is $300m$ above ground?

We know that:

\[y=0.467 \times [t^3]\]

\[300=0.467 \times [t^3]\]

\[300=0.467 \times t^3\]

\[t=8.57\ s\]

We have:

\[v_y=\int_{0}^{8.57}{2.8}{dt}\]

\[v_y=103\ m\]

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