# Mechanics 3Watch

This discussion is closed.
13 years ago
#21
(Original post by geekypoo)
Yay i've done the last part, correct me if im wrong Fermat but it uses the conservation of energy doesnt it, so:
...
The last part uses conservsation of energy, but so also does the part before, part (iii)

part (iii)
At max height, the particle still has horizontal velocity, which adds to its energy.

The energy eqn is,

where is the horizontal velocity of the particle
and is the actual velocity of the particle
and

and
m/s

Â½(1.14)Â² = 9.8h + Â½(1.14sin41.8)Â²
0.6498 = 9.8h + 0.2887
9.8h = 0.3611
h = 0.3611/9.8
h = 0.037 m
h = 3.7 cm
========

Alternatively, you can use the vertical velocity with suvat eqns

vÂ² - uÂ² = 2as
0 - (1.14cos41.8)Â² = -2gh
h = (1.14cos41.8)Â²/2g
h = 0.0368
h = 3.7 cm
========

let's hope I haven't made any more stupid mistakes!

part (iv) now becomes (using conservation of energy),

h' = 0.2 - 22/135
h' = 1/27 m
========

===========

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