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\,(\mathrm m)/(\mathrm s^2)\right)\cos50^\circ}_(a_x)\,\vec\imath+\underbrace{\left(25\,(\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)=\frac12a_xt^2\,\vec\imath+\frac12a_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\,(\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-\frac g2t^2\right)\,\vec\jmath`

The rocket will reach the ground when

`(72\,\mathrm s^2)a_y+(12\,\mathrm s)a_yt-\frac 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.)