Question

Could someone please help me with this? I've put in work but am most definitely stuck.

Answer 1

So basically I'm trying to do the first part, but am stuck.
Here's what I did:
v = v(initial) + at to get v after the engine stops
and then found the first height (when the engine works) by doing
h = vsin(angle)t + .5at^2
which translates into
h = 73 sin(56)(4) + .5(18)(4^2)
because h is concerned with the y-velocity only, right?
so I got
h = 386.07897, approx.

then, I figured that since v=at
t=vsin(angle)/a
and then plugged this in (this is the y direction)
into the 'engine stopped' leg of this problem
Stopped distance = vsin(angle)t - .5g(t^2)
and then I subbed in my value for t ( vsin(angle)/a )
and eventually got like
3v sin(angle)^2 / 2g
for the no-engine part
which I then added to the height calculated earlier.

Answer 2

It's incorrect, though. What should I do / what did I do wrong?

Answer 3

In the 1st 4 seconds (before the engine stops), there are two forces acting on the vertical component of the flight path. Gravity pulls the rocket down, while the engine's propulsion pushes the rocket up.

To find the net acceleration, you have:

g = Gravity = -9.8m/s^2
Anet = Net acceleration
Ae = Engine acceleration

Taking into account the angle at which the engine pushes:
Anet = 18m/s^2 * sin(56) - 9.8m/s^2 = 5.123m/s^2

Solving for vertical displacement in the 1st 4 seconds, with:
y = Vertical displacement
Vo = Initial velocity
t = Elapsed time
y = Vo * t + .5 * Anet * t^2
y = 73sin(56)*4 + .5 * Anet * 16
y = 73sin(56)*4 + .5 * 5.123 * 16 = 283.06m

To solve for speed at t = 4s, let Vi = Vo + Anet * t:
Vo = Velocity at 0s = 73sin56
Vi = Vo + Anet * t
Vi = 73sin56 + 5.123 * 4 = 81.01 m/s

To solve for displacement from t = 4s to maximum height, let:
Vi = Velocity at t = 4s
Vf = Velocity at maximum height = 0m/s
y = Vertical displacement
a = Acceleration = g
Vf^2 = Vi^2 + 2ay
y = - Vi^2/(2a) = 334.83m

Total vertical displacement is the sum of the two parts, or 283.06 + 334.83 = 617.89 m.