As a science fair project, you want to launch an $800g$ model rocket straight up and hit a horizontally moving target as it passes $30m$ above the launch point. The rocket engine provides a constant thrust of $15.0N$. The target is approaching at a speed of $15m/s$. At what horizontal distance between the target and the rocket should you launch?

The rocket engine provides a constant thrust of $15.0N$. The target is approaching at a speed of $15m/s$.

We analyze the situation, where the rocket moves vertically and the target moves horizontally. Let us find the time taken by the rocket from launching till it hits the target. The rocket is launched vertically due to its force $F_R=15\mathrm{N}$against its weight. So, as the net force on the rocket will be given by

$F_{net}=F_{\mathrm{R}}-m_{\mathrm{R}}g$

$m_{\mathrm{R}}{a}_{\mathrm{R}}=F_{\mathrm{R}}-m_{\mathrm{R}}g$

$a_{\mathrm{R}}=\frac{F_{\mathrm{R}}-m_{\mathrm{R}}g}{m_{\mathrm{R}}}$

Now, we plug the values for $F_{\mathrm{R}},m_{\mathrm{R}}$and $g$into equation (1) to get $a_{\mathrm{R}}$

$=\frac{15\mathrm{N}-(0.8\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{0.8\mathrm{kg}}$

$=8.95\mathrm{m}/{\mathrm{s}}^2$

The rocket starts from ground, so its initial position is $y_{i,\mathrm{R}}=0$with an initial speed $v_{i,\mathrm{R}}=0$and time $t_{i,\mathrm{R}}=0$and reaches after travelling distance $y_{f,\mathrm{R}}$in time $t_{f,\mathrm{R}}$. We need to find the time $t_{f,\mathrm{R}}$by using the value of the calculated$a_R$. The target is $30\mathrm{m}$above the launch point, so the distance that the rocket travel is $y_{f,R}=30\mathrm{m}$. From the kinematics equation, we can find the time $t_{f,R}$by

$y_{f,\mathrm{R}}=y_{i,\mathrm{R}}+{\left(v_{i,\mathrm{R}}\right)}_y\left(t_{f,\mathrm{R}}-t_{i,\mathrm{R}}\right)+\frac{1}{2}{a}_R{\left(t_{f,\mathrm{R}}-t_{i,\mathrm{R}}\right)}^2$

Now, we plug the values for $y_{f,\mathrm{R}},y_{i,\mathrm{R}},v_{i,\mathrm{R}},t_{i,\mathrm{R}}$and $a_R$into equation (2) to get $t_{f,\mathrm{R}}$

\begin{gathered}

$y_{f,\mathrm{R}}=y_{i,\mathrm{R}}+{\left(v_{i,\mathrm{R}}\right)}_y\left(t_{f,\mathrm{R}}-t_{i,\mathrm{R}}\right)+\frac{1}{2}{a}_R{\left(t_{f,\mathrm{R}}-t_{i,\mathrm{R}}\right)}^2$

$t_{f,\mathrm{R}}=2.59\mathrm{s}$

This time is taken for the rocket and the target to meet, so the target need time $t_{f,\mathrm{T}}=t_{f,\mathrm{R}}=2.59\mathrm{s}$.

The target starts from an initial position is $x_{i,\mathrm{T}}=0$with an initial speed $v_{i,\mathrm{R}}=15\mathrm{m}/\mathrm{s}$and reaches to the final point after travelling distance $x_{f,T}$in time $t_{f,T}$. The target moves with constant speed, which means its acceleration equals zero $a_T=0$. From the kinematics equation, we can find the distance $x_{f,T}$by

$x_{f,\mathrm{T}}=x_{i,\mathrm{T}}+{\left(v_{i,\mathrm{T}}\right)}_x\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)+\frac{1}{2}{a}_T{\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)}^2$

Where $x_{f,\mathrm{T}}$is the horizontal distance between the target and the rocket to be launched. Now, we plug the values for $x_{i,\mathrm{T}},v_{i,\mathrm{T}},t_{i,\mathrm{T}},t_{f,\mathrm{T}}$and $a_T$into equation (3) to get

$x_{f,\mathrm{T}}$

$x_{f,\mathrm{T}}=x_{i,\mathrm{T}}+{\left(v_{i,\mathrm{T}}\right)}_x\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)+\frac{1}{2}{a}_T{\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)}^2$

$=0+(15\mathrm{m}/\mathrm{s}{)}_x(2.59\mathrm{s}-0)+\frac{1}{2}(0){\left(t_{f,\mathrm{T}}-t_{i,\mathrm{T}}\right)}^2$

$=38.85m$

The horizontal distance between the target and the rocket should be $39\mathrm{m}$.