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Projected particle

by aurao2003
Tags: particle, projected
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aurao2003
#1
Feb17-11, 12:05 AM
P: 126
1. The problem statement, all variables and given/known data
Hi
I have been stumped by this question for the past few days. Worrying since I will be sitting STEP in June! Never mind. It goes like this:

A particle is projected vertically upwards with a speed of 30m/s from a point A. The point B is h metres above A. The particle moves freely under gravity and is above B for a time 2.4s. Calculate the value of h.




2. Relevant equations
V=U + at
V^2=U^s +2aS




3. The attempt at a solution
These are my parameters:

From A to B
u=30
a=-9.8
s=h

From B
u=?
a=-9.8
s=?
t=2.4

I took t=2.4 as the Maximum height above B.(I am not sure about this line of reasoning!)
Based on that V=0
Using V=U + at
U=23.52 m/s (This seemed to make sense since particle will start decelerating at some point)
So, U (at b)=23.52 m/s

Considering it from A to B
U=30
V=23.52
a=-9.8
S=h
Using V^2=U^s +2aS

h=17.7m (3s.f)

But alas, it seems my answer (and probable reasoning) is wrong. Any suggestions???
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TylerH
#2
Feb17-11, 12:26 AM
P: 737
You have a lot of variables and no explanation of what they are.

The way I would do this would be to solve the quadratic formula for h(t) - h, where h(t) is the function of time to hight, and h is the hight you're trying to find. h would be grouped with c. I can't desern the values of a, b, and c from your post, so you're gonna have to plug them in. Since you know that the quadratic equation is give the zeros of the function, and by subtracting h, you're shifting the function down, such that the new zeros are at the values of t are values that would have made h(t)=h
[tex]\frac{-b-\sqrt{b^2-4a(c-h)}}{2a}-\frac{-b+\sqrt{b^2-4a(c-h)}}{2a}=2.4[/tex]


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