# Projected particle

## Homework Statement

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.

V=U + at
V^2=U^s +2aS

## 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???

$$\frac{-b-\sqrt{b^2-4a(c-h)}}{2a}-\frac{-b+\sqrt{b^2-4a(c-h)}}{2a}=2.4$$