Answer on Question #49632-Physics-Mechanics-Kinematics-Dynamics
A cannon ball has a range R on a horizontal plane. If h and h' are the greatest heights in the two paths for which this is possible, then
R = 4 h h ′ . R = 4 \sqrt {h h ^ {\prime}}. R = 4 h h ′ . Solution
The two angles are complimentary.
h = u 2 sin 2 θ 2 g ; h ′ = u 2 sin 2 ( 90 − θ ) 2 g = u 2 cos 2 θ 2 g . h = \frac {u ^ {2} \sin^ {2} \theta}{2 g}; h ^ {\prime} = \frac {u ^ {2} \sin^ {2} (9 0 - \theta)}{2 g} = \frac {u ^ {2} \cos^ {2} \theta}{2 g}. h = 2 g u 2 sin 2 θ ; h ′ = 2 g u 2 sin 2 ( 90 − θ ) = 2 g u 2 cos 2 θ .
And
R = u 2 sin 2 θ g = 2 u 2 sin θ cos θ g . R = \frac {u ^ {2} \sin 2 \theta}{g} = \frac {2 u ^ {2} \sin \theta \cos \theta}{g}. R = g u 2 sin 2 θ = g 2 u 2 sin θ cos θ .
But
sin θ = 2 g h u 2 , cos θ = 2 g h ′ u 2 . \sin \theta = \sqrt {\frac {2 g h}{u ^ {2}}}, \cos \theta = \sqrt {\frac {2 g h ^ {\prime}}{u ^ {2}}}. sin θ = u 2 2 g h , cos θ = u 2 2 g h ′ .
Therefore
R = 2 u 2 g 2 g h u 2 2 g h ′ u 2 = 4 h h ′ . R = \frac {2 u ^ {2}}{g} \sqrt {\frac {2 g h}{u ^ {2}}} \sqrt {\frac {2 g h ^ {\prime}}{u ^ {2}}} = 4 \sqrt {h h ^ {\prime}}. R = g 2 u 2 u 2 2 g h u 2 2 g h ′ = 4 h h ′ .
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