Rocket Problem: Solving for Velocity, Time, and Altitude

[Libohove90]

Rocket Problem, please help :)

Homework Statement

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it runs out of fuel at an altitude of 950 meters. After this point, it's acceleration is that of gravity, downward. a) What is the velocity of the rocket when it runs out of fuel? b) How long does it take to reach this point? c) What maximum altitude does the rocket reach? d) How much time (total) does it take to reach the maximum altitude?

Homework Equations

2a x Delta Y = Vf^2 - Vi^2
a = Delta V / t
Quadratic Formula

The Attempt at a Solution

First, I need to find the velocity of the rocket when it runs out of fuel.
I used 2a x Delta Y = Vf^2 - Vi^2
Plug in: 2 (3.2 m/s^2) (950 meters) = V^2 - 0
Answer: V^2 = 6080 m^2/s^2, thus V = 78 m/s

Second, I need to find how long it takes to reach this point.
I used a = Delta V / t and solve for t which makes: t = Delta V / a
Plug in: t = (78 m/s) / (3.2 m/s^2)
Answer: 24 seconds

Third, I need to find the maximum altitude the rocket reaches. Its already at 950 m once the fuel runs out.
I used 2a x Delta Y = Vf^2 - Vi^2, where Vi = 78 m/s, Vf^2 = 0 and a = g = -9.80 m/s^2
I solve for Delta Y, thus equation is Delta Y = Vf^2 - Vi^2 / 2a
Plug in: Delta Y = (0) - (78 m/s)^2 / 2 (-9.80), which = 6084 / 19.6 = 310 m
Answer: 310 m + 950 m = 1260 m or 1.26 km

Now here's where I get confused. I need to find the total time it takes to reach that maximum altitude of 1260 m. I already calculated that it took 24 seconds to reach 950 m, before decelerating and reaching zero velocity at 1260 m.
I used Delta Y = volt + 0.5(-g)t^2
Plug in: 310 m = (78 m/s) t + (-4.90) t^2
I get: 4.9t^2 - 78t + 310 = 0
I use quadratic formula and I get 2 values for time, 7.7 and 8.2 seconds. Wtf?

Appreciate the help thank you

 

[D H]

Now here's where I get confused. I need to find the total time it takes to reach that maximum altitude of 1260 m. I already calculated that it took 24 seconds to reach 950 m, before decelerating and reaching zero velocity at 1260 m.
I used Delta Y = volt + 0.5(-g)t^2
Plug in: 310 m = (78 m/s) t + (-4.90) t^2
I get: 4.9t^2 - 78t + 310 = 0
I use quadratic formula and I get 2 values for time, 7.7 and 8.2 seconds. Wtf?

Appreciate the help thank you

You are getting two values because of rounding errors and use of an equation that is sensitive to those errors. Suppose an object is thrown upward with a velocity of 78 m/s. With g=9.8 m/s², this object would reach a maximum altitude of 310.4 meters, not 310 meters. This means it will pass through 310 meters twice, about 0.3 seconds before and 0.3 seconds after reaching the peak of 310.4 meters.

Let's see what the radical is, without reducing things to numbers. Denote

• a as the upward acceleration while the rocket is firing (3.2 m/s²)
• g as the downward acceleration due to gravity (9.8 m/s²)
• h₁ as the given 950 meters 'til rocket cutoff
• v_c as the upward velocity at rocket cutoff (your 78 m/s)
• h₂ as the additional height traveled after the rocket cutoff (your 310 meters)

You derived that [itex]v_c^2 = 2 h_1 a[/tex] and $h_2 = h_1\,a/g$. The radical in your quadratic equation is [itex]v_c^2 - 4 h_2 (g/2) = 2 h_1 a - 4 (h_1 a/g)(g/2)[/tex], and this is obviously zero. You obtained a non-zero (positive) value because you rounded some intermediate values. You are lucky in a sense; your radical could easily have been negative due to rounding errors.<br /> <br /> Perhaps you need a different equation. You are trying to find when the velocity is zero. An equation that directly addresses velocity might be useful here.[/itex][/itex]