Vertical Motion

  • Thread starter Noir
  • Start date
  • #1
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Homework Statement


A toy rocket is projected vertically upwards from rest and it rises with an acceleration of (9-5t)g m / s^2 for the first two seconds and, thereafter, freely against gravity. Find: the maximum speed and the maximum height.


Homework Equations


a = (9-5t)g for the first 2 seconds.
general motion equations

The Attempt at a Solution


The acceleration after 2 seconds is (9 - 10). -g = 9.81 m/s^2.
The velocity at this point is:
v = u + at
v = 0 + 9.81 x 2 = 19.62 m/s.
Max height occurs when v = 0 therefore ;
0 = 19.62 -9.81 x t
t is = 2, which gives a total time of 4 seconds.
I can get the the max height and velocity easily from here. Except the answers are quite bigger than what I'm getting. So I'm thinking I'm doing something wrong for these first 2 seconds. I don't think I can integrate the acceleration to find velocity because I don't have enough information.

Any help is appreciated :)

Noir
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
8
Up to 2 seconds, the acceleration is varying. So
dv = ( 9-5t)*g*dt.
To get the velocity at 2 seconds,integrate dv between 0 to 2 seconds. Then proceed.
 
  • #3
Delphi51
Homework Helper
3,407
11
v = 0 + 9.81 x 2 = 19.62 m/s.
It looks like you are using the formula Vf = Vi + a*t here.
But that formula only applies when the acceleration is constant.
Your acceleration varies with time during the first two seconds.
You can deal with varying acceleration by using calculus (integration) or by drawing an acceleration vs time graph using your knowledge that the area under the graph is the velocity.
 

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