Howard J. answered • 09/23/19

Down-to-Earth Engineer for Math, Engrg, Drafting, and Physics Help

OK, so the rocket takes off from the ground (elevation=0) at time t=0 with a constant acceleration of 7.5 m/sec^{2} and continues for 100 seconds. At that time its acceleration goes from 7.5 m/sec^{2} up to g=9.81 m/sec^{2} down. That’s when this becomes a simple vertical projectile problem. So, after the 100 sec is up, the problem is just like a projectile motion problem but in only one dimension. Here’s how this thing goes. I’m going to assume that you have no calculus background and can pull the equations from memory. I’m going to be thorough. Normally such a problem can take much less space.

Given: Over 0 ≤ t ≤ 100 sec, a=7.5 m/sec^{2} upward starting from the ground.

Find: The total height reached.

Let E(t)=elevation as function of time.

For constant acceleration motion in one dimension (1D),

E(t)=E(0)+v(0)t+0.5at^{2}. At t=0, E(t)=0. In other words, E(0)=0. When I write E(0) that doesn’t mean E times 0. It means E(t) when t=0. Also, since the rocket starts from a standstill, the initial velocity is zero v(0)=0. This simplifies the equation:

E(t)=0+0+0.5at^{2}=0.5at^{2}. We are given the acceleration, a, so,

E(t)=0.5(7.5)t^{2}=3.75t^{2}. Now when t=100 sec, we can calculate the elevation E(100)=3.75(100)^{2}=37,500 m

From now on let's work in km rather than m.

At this time the acceleration changes from a=+7.5 m/sec^{2} to a=-g=-9.81 m/sec^{2} =-.00981 km/sec^{2} and start with a new elevation equation:

For t≥100 sec, E(t)=E(100)+v(100)t-0.5gt^{2}.

We know E(100) but we don’t know v(100). So, we have to go back and figure that out. For constant acceleration v(t)=v(0)+at. Since v(0)=0,

v(t)=at=(7.5/1000)t=0.0075t km/sec

v(100)=(0.0075)(100)= 0.75 km/sec (that thing’s *really moving*!).

So now, for t ≥ 100 sec,

E(t)=E(100)+0.75t-0.5(0.00981)t^{2}=37.5+0.75t -0.0049t^{2} km

When the velocity become zero, the rocket is momentarily at rest at the peak of its trajectory. So we need the equation for velocity. This is v(t)=v(100)-gt and we set this to 0, and get 0=0.75-0.00981t giving t=0.75/0.00981= 76.4 sec.

The additional height reached is then 0.75(76.4)-0.0049(76.4)^{2}=28.7 km

So the total height reached is 37.5+28.7= 66.2 km

Since the problem statement has two significant figures, this rounds to 66 km=66,000 m

Solution: 66 km