Trajectory of a Projectile on a Hill
Trajectory of a Projectile on a Hill formulas |
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\( t = ( 2 \; v_{0h} \;/\; g ) \pm \sqrt{ ( v_{0h}^2 \;/\; g^2 ) - ( 2 \;h \;/\; g ) } \) \( d = v_{0d} \; t \) \( h = v_{0h} \; t - \frac{1}{2} \; g \; t^2 \) |
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| Symbol | English | Metric |
| \( t \) = time | \(sec\) | \(s\) |
| \( g \) = gravitational acceleration | \(ft\;/\;sec^2\) | \(m\;/\;s^2\) |
| \( v_0 \) = launch velocity | \(ft\;/\;sec\) | \(m\;/\;s\) |
| \( \theta \) = vertical angle | \(rad\) | \(rad\) |
Tags: Velocity Projectile