perhitungan jarak pada gerak parabola di bidang miring

h=d ∙sinα

x=d ∙cosα

v0x=vx=v0∙cos (α+β )

v0 y=v0 ∙ sin (α+β )

02=v0 y2 −2 ∙ g ∙ hmax

2 ∙ g ∙ hmax=v02 ∙ sin2 (α+β )

hmax=v02 ∙ sin2 (α+β )2 ∙ g

∆ h=hmax−h

∆ h=v02 ∙ sin2 (α+β )2 ∙ g

−d ∙ sinα

∆ h=v02 ∙ sin2 (α+β )−2 ∙ d ∙ g ∙sinα

2 ∙ g

0=v0 y−g ∙ t 1

g ∙t 1=v0 ∙ sin (α+ β )

t 1=v0 ∙sin (α+β )

g

∆ h=12∙ g ∙ t 2

2

t 22=2∙∆h

g

t 2=√ 2( v02 ∙ sin2 (α+β )−2∙ d ∙ g ∙ sinα

2∙ g )g

t 2=√ v02 ∙ sin2 (α+ β )−2∙ d ∙ g ∙ sinαg2

x=v0x ∙ t x

t x=xv0 x

t x=d ∙cosα

v0 ∙cos (α+β )

t x=t 1+t2

d ∙cosαv0 ∙cos (α+ β )

=v0 ∙ sin (α+β )

g+√ v02 ∙ sin2 (α+β )−2∙ d ∙ g ∙ sinα

g2

d ∙g ∙cosαv0 ∙cos (α+ β )

=v0 ∙ sin (α+β )+√v02 ∙ sin2 (α+ β )−2∙ d ∙ g ∙ sinα

( d ∙g ∙cos αv0 ∙cos (α+ β )−v0 ∙sin (α+β ))

2

=v02 ∙sin2 (α+β )−2 ∙ d ∙ g ∙ sinα

( d ∙g ∙cosα−v02 ∙ sin (α+ β ) ∙cos (α+ β )

v0 ∙cos (α+β ) )2

=v 02∙ sin

2 (α+β )−2 ∙ d ∙ g ∙sin α

d2 ∙ g2∙cos2α−2 ∙ v02 ∙ d ∙ g ∙cos α ∙ sin (α+β ) ∙cos (α+β )+v0

4 ∙sin2 (α+β ) ∙cos2 (α+β )v02 ∙cos2 (α+ β )

=v02 ∙sin2 (α+β )−2 ∙ d ∙ g ∙ sinα

d2 ∙ g2 ∙cos2α−2∙ v02 ∙ d ∙ g ∙cosα ∙ sin (α+ β ) ∙cos (α+β )+v0

4 ∙ sin2 (α+β ) ∙cos2 (α+β )=v04 ∙ sin2 (α+β ) ∙cos2 (α+β )−2 ∙ v0

2 ∙ d ∙ g ∙ sinα ∙cos2 (α+β )

d2 ∙ g2 ∙cos2α−2∙ v02 ∙ d ∙ g ∙cosα ∙ sin (α+ β ) ∙cos (α+β )=−2 ∙ v0

2∙ d ∙ g ∙ sinα ∙cos2 (α+ β )

d ∙g ∙cos2α−2 ∙ v02∙cos α ∙ sin (α+β ) ∙cos (α+β )=−2∙ v0

2 ∙ sinα ∙cos2 (α+ β )

d ∙g ∙cos2α=−2 ∙ v02 ∙sinα ∙cos2 (α+β )+2 ∙ v0

2∙cos α ∙sin (α+β ) ∙cos (α+β )

d=2 ∙ v0

2 ∙cos (α+β ) (−sinα ∙cos (α+β )+cos α ∙sin (α+β ) )g ∙cos2α

d=2 ∙ v0

2 ∙cos (α+β ) [−sinα (cos α ∙cos β−sinα ∙ sin β )+cosα (sinα ∙cos β+sin β ∙cosα ) ]g ∙cos2α

d=2 ∙ v0

2 ∙cos (α+β ) (−sinα ∙cos α ∙cos β+sin2α ∙sin β+sinα ∙cos α ∙cos β+sin β ∙cos2α )g∙cos2α

d=2 ∙ v0

2 ∙cos (α+β ) (sin2α ∙ sin β+sinβ ∙cos2α )g ∙cos2α

d=2 ∙ v0

2 ∙cos (α+β ) ∙sin β (sin2α+cos2α )g ∙cos2α

d=2 ∙ v0

2 ∙cos (α+β ) ∙sin βg ∙cos2α