How Do You Calculate Maximum Velocity and Height for a Rocket Launch?

[Jimmy Rose]

Ok I've been trying to do a problem and I can't seem to get it.
A rocket is launched from the origin with an acceleration of 20m/s^2 at an angle of 30.0 degrees above the horizontal. The launch acceleration lasts for 2.00 seconds at which time the fuel is exhausted. The rocket then falls with an acceleration of 9.80m/s^2 downward.
-Ignoring air resistance.

What is the maximum velocity in the upward direction?
I used: final velocity = initial velocity + acceleration * time
Which putting in the values I got: Vf = 0 + (20(sin30))(2) = 20m/s
I also have another way
Vf = 20(sin30) + (9.8)(2) = 29.6m/s
I don't know which one is right if either
Here are the answers selections: (a) 20.0 (b) 25.0 (c) 30.0 (d) 35.0 (e) 40.0
He might or might have not rounded them.

The next question I can't even begin
What is the maximum height?
I use the equation y = Vi + 1/2at^2 and I don't get the correct answer
Here are the answers selections: (a) 22.3 (b) 27.5 (c) 30.5 (d) 36.7 (e) 40.4

I’m so frustrated I want to bit my head off especially since its multiple choices.
Please in need of Help.

 

[atyy]

I believe that at least one of your two answers for part one is right.

For part two, first calculate the height reached from the start of its motion until the time its fuel is used up (ie. the time it attains the velocity you calculated in part one). Let this height be H1.

The rocket does not start falling the moment its fuel is used up. From that moment, it will have a constant downward acceleration of 9.8 m/s², but it continues to go up because it has an upward velocity at that point. It will slow down, because of the downward acceleration, and after some additional time will its vertical velocity be 0 m/s. Only then will it start to move downward.

So first you need to find the additional time it takes for its vertical velocity to reach 0 m/s, then plug that time into the formula y = Vi + 1/2at^2. Let this be height be H2.

Finally, add H1 and H2.

 

[Sreeja Mobin]

Jimmy
The equation you used is correct. But the values you subsituted are incorrect. We can use the equation v= u+ at where v-final velocity,u-initial velocity,a-acceleration. Here final velocity is zero because after reaching the maximum height the velocity becomes zero and then started to fall downwards.
So 0= u + 20*2
i.e., u = 40m/s
the maximum height h=u*u*sin30*sin30/2*g
i.e., h=40*40*sin30*sin30/2*9.8 = 20.4m

 

[Jimmy Rose]

Okay I got 20 meters from H1 using y = 1/2 (Vf + Vi)t
y = 1/2 (20m/s + 0) 2 = 20m/s
But how would you suggest in getting the second Height.
Would this be right: Vi = 20m/s, Vf = 0, a = 9.8, y=?.
First need time before I could use y = Vit + 1/2 at^2
Vf = Vi +at
0 = 20 + 9.8t
This is where I'm stuck because how do get time from here.
It would have to be around -1.03. That doesn't really make sense to me.

 

[Jimmy Rose]

Sreeja Mobin:
Thanks but how did you get 40m/s.
And for the second equation are all the selected answers incorrect because if that's true I really didn't stand a chance lol.

 

[atyy]

But how would you suggest in getting the second Height.
Would this be right: Vi = 20m/s, Vf = 0, a = 9.8, y=?.
First need time before I could use y = Vit + 1/2 at^2
Vf = Vi +at
0 = 20 + 9.8t
This is where I'm stuck because how do get time from here.

In part 1, you used "final velocity = initial velocity + acceleration * time" to compute the final velocity given an initial velocity, an acceleration and a time. So you can use that same formula to compute a time given a new initial velocity (ie. the final velocity you computed in part one), the downward acceleration 9.8 m/s², and the final velocity 0 at the top of its trajectory.

 

[Jimmy Rose]

In part 1, you used "final velocity = initial velocity + acceleration * time" to compute the final velocity given an initial velocity, an acceleration and a time. So you can use that same formula to compute a time given a new initial velocity (ie. the final velocity you computed in part one), the downward acceleration 9.8 m/s², and the final velocity 0 at the top of its trajectory.

I'm sorry but terribly confused.
What is time though because to use "final velocity = initial velocity + acceleration * time" to find initial velocity you need time.
Because I tried to find time but it came out weird like -1.03 when I used
Vf = Vi +at
0 = 20 + 9.8t

 

[atyy]

I'm sorry but terribly confused.
What is time though because to use "final velocity = initial velocity + acceleration * time" to find initial velocity you need time.
Because I tried to find time but it came out weird like -1.03 when I used
Vf = Vi +at
0 = 20 + 9.8t

What would you get if you considered the acceleration to be in the opposite direction as your new initial velocity?

 

[Jimmy Rose]

What would you get if you considered the acceleration to be in the opposite direction as your new initial velocity?

Time would be 1

Vf = Vi +at
0 = -9.8 + 9.8t
Is that what you meant?
Because I would get a negative in Max height if Vi was downward acceleration.

 

[atyy]

Vf = Vi +at
0 = 20 + 9.8t

I was thinking that 20 is upwards, but gravity is downwards.

Vi=20
a=-9.8

 

[Jimmy Rose]

Yea!
Vf = Vi + at
0 = 20 + -9.8t
t = 2.04s

y2 = Vit + 1/2at^2
y2 = 20(2.04) + 1/2(-9.8)(2.04)^2
y2 = 20.4m

y1(20) + y2(20.4) = yfinal (40.4m)
and the third question is time it took to get there?
Which the initial 2 plus the additional 2.04 = 4.04sec which is on the answer selection.
Thanks just one more thing if not too much trouble.
What is the X distance (range) when the rocket hits the ground?
(a) 153m (b) 166m (c) 187m (d) 205m (e) 230m
All I could come up with is Vi = 0, ax=0m/s^2

 

[atyy]

Yea!

Well, glad you're happy. BTW, hope it makes sense to you, since it certainly doesn't to me - I'm a biologist and haven't done these problems in years.

Thanks just one more thing if not too much trouble.
What is the X distance (range) when the rocket hits the ground?
(a) 153m (b) 166m (c) 187m (d) 205m (e) 230m
All I could come up with is Vi = 0, ax=0m/s^2

In your solution to part one, you got the vertical component of the acceleration. You could also consider the horizontal component of the acceleration. Again, remember that there are two parts to the trajectory, each with a different acceleration.