Max Altitude Reached by Rocket at 53° Angle

Click For Summary

Homework Help Overview

The problem involves a rocket launched at a 53-degree angle with an initial speed of 100 m/s, experiencing acceleration for 3 seconds before transitioning to projectile motion. The goal is to determine the maximum altitude reached by the rocket.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of velocity components and the effects of acceleration on the rocket's motion. There is uncertainty regarding which initial vertical velocity to use when calculating the final velocity after the acceleration phase.

Discussion Status

Some participants have provided guidance on breaking the problem into two parts: the initial phase with constant acceleration and the subsequent projectile motion. There is ongoing exploration of how to handle the transition between these phases, with no explicit consensus reached.

Contextual Notes

Participants are considering the implications of the acceleration phase and the transition to free-fall conditions, as well as the assumptions regarding the rocket's motion during these phases.

Toranc3
Messages
189
Reaction score
0

Homework Statement



a rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. the rocket moves for 3 s along its initial straight line of motion with an acceleration of 30m/s^2. FInd the max altitude reached by the rocket

Homework Equations



y=y0 + v0*t +1/2*g*t^(2)

v0y=v0sin(53)
v0x=v0cos(53)

vy=v0y+g*t


The Attempt at a Solution



velocity components:

v0y=100m/s*sin(53)=79.86m/s
v0x=100m/s*cos(53)=60.18m/s

Found the vertical distance with the acceleration of 30m/s^(2) for 3sec

y=y0 + v0*t +1/2*g*t^(2)
y=79.86m/s*3 sec + 1/2*30m/s*(3sec)^(2) =374.58m

now I am trying to find my final velocity for the acceleration of 30m/s^(2) for 3sec because that will be my initial velocity for the acceleration of 9.8m/s^(2)

I use this formula, vy=v0y+gt

I am not sure which v0y do I use. Do i just use the initial velocity of 100m/s or do I use my v0y that I calculated?(100*sin(53)=79.86m/s) Which one do I use and why? thanks
 
Physics news on Phys.org
a rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. the rocket moves for 3 s along its initial straight line of motion with an acceleration of 30m/s^2. FInd the max altitude reached by the rocket

----------------------------------------
I guess after 3sec. the power is turned off.
Calculate(v1,x,y) first part of the flight given - v0=100m/s, θ=53° , 3sec and 30m/s2.
Second part is projectile motion.
 
azizlwl said:
a rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. the rocket moves for 3 s along its initial straight line of motion with an acceleration of 30m/s^2. FInd the max altitude reached by the rocket

----------------------------------------
I guess after 3sec. the power is turned off.
Calculate(v1,x,y) first part of the flight given - v0=100m/s, θ=53° , 3sec and 30m/s2.
Second part is projectile motion.

I see but how come I can't break up the first part into components?
 
First part is just a straight line with constant acceleration.
 
azizlwl said:
First part is just a straight line with constant acceleration.

Ok thank you. :)
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 16 ·
Replies
16
Views
7K