Calculating Time to Reach Ground from Rocket Launch

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In summary, the problem involves a rocket starting from rest and accelerating upwards for 3 seconds at 30 m/s^2, then accelerating downwards at -6 m/s^2 for an unknown time until it reaches the ground. By breaking the problem into two stages and using equations for distance and time, the total time is found to be approximately 34 seconds.
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Homework Statement


How much time does it take from the rocket gets shot up, till it reaches the ground again?
The rocket accelerates 30m/s^2 for the first 3 seconds, and accelerates downwards -6m/s^2 after.

Homework Equations


(v-v0) / a = t to find time

The Attempt at a Solution


2ad=v^2-v0^2

d= time
v=speed before landing
v0= starting speed 45m/s

I found v0 to be 45, (810/18)
810m the highest it gets, after 18 seconds.

i find v0^2 to be 2025

the distance must be 810 times 2? so 1620meter

I know the acceleration is -6 but ill just write 6 here for now
2*(6)*1620= 19440

2025+19440=21465

so v^2=21465

v=146

i go back to v=v0+at

(v-v0) / a = t

(146 - 45) / 6 = 16

the answer is supposed to be 34sec, and I am confused:)
my guess is i do something wrong with v or v0, i take v0 as the starting point (at 18 sec)
 

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  • #2
You have found correctly that it takes 18 s for the rocket to reach a maximum height of 810 m. What is the instantaneous speed of the rocket at this max. height? Given this initial speed, how long will it take for it to come back down? Add this to the 18 s.
 
  • #3
You will need to break the problem into two segments since the rocket has different stages of acceleration and you're quoting formulas which assume constant acceleration.

Question 1: Where is the rocket and how fast is it moving after the first 3 seconds of upward acceleration, given it starts at rest on the ground.
Answer 1: x = Bla, v = Bla-Bla

Question 2: How long does it take a rocket with initial position x=Bla and v = Bla-Bla to stop ascending and fall back down to a height of x=0?
Answer 2: t = Bla-Bla-Bla

Final answer to the whole question: 3s + Bla-Bla-Bla since we include both times together.

In both stages of the relevant equations will be:
##(v-v_0) = a(t-t_0)## and ##(x-x_0) = v_0(t-t_0) + \frac{1}{2}a(t-t_0)^2##, where ##x## is the position and ##v## the velocity at time ##t## and where ##x_0, v_0## is the position and velocity at #t_0# and #a# is the constant acceleration. Pick ##x=0## as the ground and fire away.
 
  • #4
kuruman said:
You have found correctly that it takes 18 s for the rocket to reach a maximum height of 810 m. What is the instantaneous speed of the rocket at this max. height? Given this initial speed, how long will it take for it to come back down? Add this to the 18 s.

The speed is 0, i figured after reading your comment
im not sure what i do wrong, when v0=0

2ad=v^2-v0^2
2*6*810=v^2-v0^2

im only getting v to be 98..and when i do
(v-v0) / a = t (right?)
i end up with 98/6 which is wrong..
 
  • #5
Joakim 1 said:
2ad=v^2-v0^2
2*6*810=v^2-v0^2
You are using the wrong equation for the job. You need an equation that involves time and distance. Solve this problem: A rocket starts from rest at 810 m and drops down with an acceleration of 6 m/s2. How long does it take to hit the ground?
 
  • #6
kuruman said:
You are using the wrong equation for the job. You need an equation that involves time and distance. Solve this problem: A rocket starts from rest at 810 m and drops down with an acceleration of 6 m/s2. How long does it take to hit the ground?

2ad=v^2-v0^2
2*6*810=v^2-v0^2

v^2=9720
v=98

v=v0+at
t=(v-v0)/a
t= 98/6 = 16sec

16 +18=34sec

is this wrong?
 
  • #7
Joakim 1 said:
16 +18=34sec

is this wrong?
Looks about right, but could be more accurate.
As @jambaugh noted, there is a better way to subdivide the problem. For the purposes of the question, there is nothing special about reaching maximum height; what is special is the change in acceleration.

So we have first stage
a=30m/s2
t=3s
vi=0
vf=90m/s
s=vit+½at2

Second stage
a=-6m/s2
s=-s from first stage
vi=vf from first stage.
s=vit+½at2

Solve to find t, being careful to select the correct root.
 

1. How is the time to reach ground from a rocket launch calculated?

The time to reach ground from a rocket launch is calculated by using the equation t = √(2h/g), where t is the time in seconds, h is the initial height of the rocket in meters, and g is the acceleration due to gravity (9.8 m/s^2). This equation only applies to objects in freefall, so it is important to ensure that the rocket is not experiencing any external forces such as air resistance.

2. What factors affect the time to reach ground from a rocket launch?

The time to reach ground from a rocket launch is affected by the initial height of the rocket, the acceleration due to gravity, and any external forces such as air resistance or wind. The shape and weight of the rocket can also have an impact on the time it takes to reach the ground.

3. Can the time to reach ground from a rocket launch be calculated for different planets?

Yes, the time to reach ground from a rocket launch can be calculated for different planets using the same formula t = √(2h/g). However, the value of g will be different for each planet depending on its mass and radius. For example, on Mars where the acceleration due to gravity is 3.71 m/s^2, the time to reach ground from the same initial height will be longer compared to Earth.

4. Is the time to reach ground from a rocket launch affected by the angle of launch?

Yes, the time to reach ground from a rocket launch is affected by the angle of launch. The equation t = √(2h/g) assumes that the rocket is launched vertically, so a different angle will result in a different time. If the rocket is launched at an angle, the horizontal and vertical components of its velocity must be taken into account to calculate the time to reach ground.

5. How accurate is the calculation of time to reach ground from a rocket launch?

The calculation of time to reach ground from a rocket launch is relatively accurate under ideal conditions, assuming no external forces are acting on the rocket. However, in real-life scenarios, there may be factors such as air resistance, wind, and variations in the acceleration due to gravity that can affect the accuracy of the calculation. Additionally, the shape and weight of the rocket may also impact the time it takes to reach the ground. Therefore, the calculated time should be considered an estimate rather than an exact value.

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