# Projected particle

• Feb 16th 2011, 09:01 PM
gbenguse78
Projected particle
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.
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???
• Feb 16th 2011, 09:15 PM
alexmahone
Quote:

Originally Posted by gbenguse78
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.
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???

The particle moves up from B and returns to B. So the displacement is 0.

$\displaystyle s=v_Bt-\frac{1}{2}gt^2$

$\displaystyle 0=v_B(2.4)-\frac{1}{2}(9.8)(2.4)^2$

$\displaystyle v_B=11.76\ m/s$

$\displaystyle v_B^2=v_A^2-2gh$

$\displaystyle 11.76^2=30^2-2(9.8)h$

$\displaystyle h=38.8624\ m$
• Feb 17th 2011, 12:50 AM
gbenguse78
Quote:

Originally Posted by alexmahone
The particle moves up from B and returns to B. So the displacement is 0.

$\displaystyle s=v_Bt-\frac{1}{2}gt^2$

$\displaystyle 0=v_B(2.4)-\frac{1}{2}(9.8)(2.4)^2$

$\displaystyle v_B=11.76\ m/s$

$\displaystyle v_B^2=v_A^2-2gh$

$\displaystyle 11.76^2=30^2-2(9.8)h$

$\displaystyle h=38.8624\ m$

Thanks. Much appreciated!