DC Mathematica 2018

𝑆 = 𝑉  cos cos

from equation ①

 =
cos
Vo cos α

from equation ①

− 𝑉  cos sin
cos

= 𝑉  sin − 1 2

 2 substituting 𝑆 = 𝑉  cos cos

into equation ②

−𝑉 cos tan
= 𝑉 sin − 1 2

 using sin
/cos
= tan
and dividing by t

sin α − 1 2

g s cos B V 𝑜 cos α

substituting in  =
cos
Vo cos α

−𝑉 cos α tan β = V o

2𝑉 2 cos 2 tan
+2𝑉 2 sin cos = 
cos

Rearranging the equation

𝑆 = 2𝑉 2 cos 2 
+2𝑉 2 sin cos  cos

Putting the equation in term of So

𝑆 = 2𝑉 2 cos  cos

(sin + cos tan
) Rearranging and factorising

𝑆 = 2𝑉 2 cos 2  cos

( sin+cos tan
cos )

multiplying by cos cos

𝑆 = 2𝑉 2 cos 2  cos

(tan + tan
)

using sin /cos = tan , and simplifying

𝑆 = 2𝑉 2 cos 2 (tan+tan
)  cos

③

Then, we have to find an equation for t , airborne time, which allows the snowboarders to perform the most tricks.

= 2𝑉 2 cos 2 (tan+tan
)  cos

𝑉  cos cos

substituting in 𝑆 = 𝑉  cos cos

(from①)

𝑉  cos = 2𝑉 2 cos 2 

(tan+tan
)  cos

multiplying both sides by cos

 = 2𝑉 2 cos 2 

(tan+tan
)  cos
𝑉 cos

dividing both sides by 𝑉 cos

 = 2𝑉 
(tan+tan
) 

simplifying, for equation ④

We then need to find the equation for the maximum height which can be achieved, since this is often criteria for marking in the snowboard halfpipe, the height which is shown in the diagram below as H max .

V = 0

V v

V o

H max .



u = Vv

From SUVAT,  2 =  2 + 2
, hence
=  2 − 2 2

, where  = 0,  = 𝑉 = 𝑉

2 and = −, we

can find that:

𝑉 2 sin 2 2

−𝑉 −2

𝐻 =

=

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