Question

Can someone solve this for me? Thanks

Answer 1

I think it's safe to say the rocket starts at rest; let its starting position serve as the origin. We're given the rocket's initial acceleration vector, and presumably while the engines are firing, this acceleration is constant up until burn-out, at which point the rocket will only experience a downward acceleration due to gravity.

Translating the given info and the above into vectors with components, we have for the first 12 seconds of the engines firing,

$\vec a_{0\le t\le12\,\mathrm s}=\underbrace{\left(25\,\dfrac{\mathrm m}{\mathrm s^2}\right)\cos50^\circ}_{a_x}\,\vec\imath+\underbrace{\left(25\,\dfrac{\mathrm m}{\mathrm s^2}\right)\sin50^\circ}_{a_y}\,\vec\jmath$

$\vec v_{t=0}=\underbrace{0}_{v_x}\,\vec\imath+\underbrace{0}_{v_y}\,\vec\jmath$

$\vec r_{t=0}=0\,\vec\imath+0\,\vec\jmath$

We integrate $\vec a$ twice to find a solution for $\vec r$:

$\vec v_{0\le t\le12\,\mathrm s}=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau$

$\vec v_{0\le t\le12\,\mathrm s}=\left(\displaystyle\int_0^ta_x\,\mathrm d\tau\right)\,\vec\imath+\left(\displaystyle\int_0^ta_y\,\mathrm d\tau\right)\,\vec\jmath$

$\vec v_{0\le t\le12\,\mathrm s}=a_xt\,\vec\imath+a_yt\,\vec\jmath$

$\vec r_{0\le t\le12\,\mathrm s}=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau$

$\vec r_{0\le t\le12\,\mathrm s}=\left(\displaystyle\int_0^tv_x\,\mathrm d\tau\right)\,\vec\imath+\left(\displaystyle\int_0^tv_y\,\mathrm d\tau\right)\,\vec\jmath$

$\vec r_{0\le t\le12\,\mathrm s}=\dfrac12a_xt^2\,\vec\imath+\dfrac12a_yt^2\,\vec\jmath$

So at the end of the first 12 second interval, we have velocity and position vectors, respectively,

$\vec v_{t=12\,\mathrm s}=(12\,\mathrm s)a_x\,\vec\imath+(12\,\mathrm s)a_y\,\vec\jmath$

$\vec r_{t=12\,\mathrm s}=(72\,\mathrm s^2)a_x\,\vec\imath+(72\,\mathrm s^2)a_y\,\vec\jmath$

After the engine burns out, the rocket has a constant accleration vector

$\vec a_{t>12\,\mathrm s}=-g\,\vec\jmath$

where $g=9.8\,\frac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. Using the velocity and position at the 12 second mark, we integrate this vector twice to find the velocity and position vectors for any time $t$ after 12 seconds:

$\vec v_{t>12\,\mathrm s}=\vec v_{t=12\,\mathrm s}+\displaystyle\int_{12}^t\vec a_{t>12\,\mathrm s}\,\mathrm d\tau$

$\vec v_{t>12\,\mathrm s}=(12\,\mathrm s)a_x\,\vec\imath+\left((12\,\mathrm s)a_y-gt\right)\,\vec\jmath$

$\vec r_{t>12\,\mathrm s}=\vec r_{t=12\,\mathrm s}+\displaystyle\int_{12}^t\vec v_{t>12\,\mathrm s}\,\mathrm d\tau$

$\vec r_{t>12\,\mathrm s}=\left((72\,\mathrm s^2)a_x+(12\,\mathrm s)a_xt\right)\,\vec\imath+\left((72\,\mathrm s^2)a_y+(12\,\mathrm s)a_yt-\dfrac g2t^2\right)\,\vec\jmath$

The rocket will reach the ground when

$(72\,\mathrm s^2)a_y+(12\,\mathrm s)a_yt-\dfrac g2t^2=0\implies t=52\,\mathrm s$

This is the total time the rocket spends in flight (taken here to mean the entire time it's in the air, not just after the engines burn out - if you want to omit the initial launch period, just subtract 12 seconds).

The rocket's horizontal displacement after this time is

$(72\,\mathrm s^2)a_x+(12\,\mathrm s)a_x(52\,\mathrm s)=11,000\,\mathrm m$

Its maximum height occurs when the vertical component of the rocket's velocity vector is 0, at which point we use the equation

${v_f}^2-{v_i}^2=-2g(y_f-y_i)$

where in this case we're interested in times after the 12 second mark, and use the velocity/position at 12 seconds as initial conditions. With $v_f=0$, we would have $y_f=y_{\mathrm{max}}$, so we end up with

$-\left((12\,\mathrm s)a_y\right)^2=-2g(y_{\mathrm{max}}-(72\,\mathrm s^2)a_y)$

$\implies y_{\mathrm{max}}=4100\,\mathrm m$

(Note that all answers were rounded to two significant digits.)