Find the maximum altitude reached by the rocket.?

Question

A rocket is launched at an angle of 55.0° above the horizontal with an initial speed of 98 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile.

(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m

Answer 1

initial displacement = dboost = v0t + 1/2at^2
hboost = dboost sin theta
xboost = dboost cos theta
vboost = v0 + at
vxboost = vboost cos theta
vyboost = vboost sin theta

To get the maximum altitude, use conservation of energy
initial PE + vertical KE
= mg (hboost) + 1/2 m vyboost^2
= final PE = mg (hfinal)

hfinal = hboost + vyboost^2 / 2g
= hboost + ((v0 + at) * sin (theta))^2 / 2g

b) flight time =
boost time + time to rise ballistically from hboost to hfinal
+ time to fall from hfinal

The time to fall a distance h can be gotten by:
h = 1/2 gt^2, so t = sqrt (h/2g)

So flight time = t + sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g)

Horizontal range = xboost + vxboost * time of ballistic flight

= (v0t + 1/2at^2) cos theta
+ (v0 + at) cos theta * (sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g))

Lots of plugging and chugging to do now.

EDIT--the problem said acceleration was constant, so we don't have to worry about loss of mass during the boost. Realistic? No. But that's what the problem says. Once the thing is ballistic, it's a moot point anyway.

And hboost is given, but dboost and an angle are, so getting hboost is just simple trig (as I did above)

Answer 2

The conservation of energy law tells us that total energy TE will be preserved no matter where in the trajectory the rocket is.

At flame out TE = PE + KE + W; where PE = mgh for the empty rocket mass m at flame out height h above ground. KE = 1/2 mV^2; where V^2 = U^2 + 2aD = (U + at)^2 and U = 98 mps, t = 3 sec, and a = 30 m/sec^2. V = U + at = 98 + 30*3 = 188 mps, the burn out end velocity.

Work on the rocket to convert potential energy from the fuel to kinetic energy is W = FD = MaD; where a = 30 m/sec^2 and D is the distance traveled while boosting. From above 2aD = V^2 - U^2; so that D = [V^2 - U^2]/2a

Unfortunately, M = m + f; where m is the empty rocket fixed mass and f is the fuel variable mass. And we know that f > m in many if not most cases. So it cannot be ignored. And, as fuel mass was not given, work expended to propel the rocket to V burn out velocity cannot be properly assessed. We can only say that it will be less than W = MaD because f = M - m and average fuel, if there is even burning over time, would be f/2; so that M(f) = m + f/2 < m + f = M.

[NB: I'm going to ignore the fact that the overall mass of the rocket is diminishing during boost phase. Otherwise, the calculations get really messy. But in reality the accelerating force F = d(mv)/dt = dm/dt v + m dv/dt does the work W and, as you can see, dm/dt v is the mass reduction term in that force.]

Putting this all together, we have TE = mgh + 1/2 mV^2 + m[V^2 - U^2]/2 = mgh + mV^2 - 1/2 mU^2 = m[gh + V^2 - U^2/2], and the values for all the factors on the RHS are either given or derived. So we can find the TE of the system, in this case, the rocket at flame out. That will be the same TE on impact after hitting apex, and the same TE at apex.

For example, at apex TE = PE + KE + W = m[gh + V^2 - U^2/2]; where PE = mgH, KE = 1/2 mv^2, and W = mgS. H is the total height above ground, v = vh = U cos(55) = constant, the horizontal velocity because its the only source of KE as vv, the vertical velocity, = 0 at apex. W, the work, is just W = mgS, where H = h + S and S is the vertical height measured from h and v^2 = u^2 - 2gS; where u = V at flame out.

Again, putting all this together in a balance sheet (equation), we have TE = m[gh + V^2 - U^2/2] = mgH + 1/2 mv^2 + mgS; so that gh + V^2 - U^2/2 = gH + (U cos(55))^2/2 + g(H - h). At this point, we need to know h to find H. That turns out to be h = D sin(55) and we found D earlier. So there you have it. Solve the energy balance equation above for H.

Bottom line, these kinds of problem can be best solved by recognizing TE = constant throughout the trajectory. Then all you need to do is decide what goes into PE, KE, and W at each point. [NB: check my algebra...even if there are mistakes, the approach is valid]

PS: Bekki's solution also has the same issues...mass during boost is not m, but M = m + f(dm/dt,t); where f(dm/dt,t) depends on the rate (dm/dt) and time (t) of burn. Also, she assumes some PE = mgh(boost), but no h(boost) was given...we don't know how high the rocket was off the ground when initially fired. In fact, the real potential energy at ground level is just the potential energy found in the fuel. When it burns, its potential energy is converted to kinetic energy and other forms of energy, like work and mgh.

Answer 3

You exceeded over the value of the acceleration, so i'm going to basically call it Z: the rate at 10 s is V36 = 80+10*Z. Vxo = V36*cos36° Vyo = V36*sin36° the place o is the begining of the ballistic trajectory. Xo = Vxo*10 + ½*10²*Z*cos36° = 10Vxo + 40.40 5*Z Yo = Xo*tan36° you are able to determine the rest from the eqs of ballistic action......