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say ramp angle = 30deg from horizontal, launch velocity = 10 m/s, launch point is 1m above floor

break launch velocity vector into horizontal and vertical components

horizontal = cos 30 * 10 = 8.66 m/s, vertical = sin 30 * 10 = 0.5*10=5 m/s

the horizontal component is constant so if you multiply the time the ball spends going up and down by 8.66 this represents horizontal travel

take g as -10 m/s^2, u=5, v=0 ,s=?, t=?

find height reached and time taken, t=v-u/g = 0.5secs, s=(u*t)+((g*t^2)/2)=1.25m, now add how long to fall 1.25 m+ original height (1m) = 2.25m

remember g is now +10 m/s^2, u=0, s=2.25m ,t=?

then t= sq root(s*2/10)=0.671secs(falling), add to 0.5 secs(rising) = 1.171secs, multiply by 8.66m/s then horizontal travel = 10.14 metres

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