using the usual equation for the range of a missile,
R = v^2/g sin2θ
we get
427^2/9.81 sin2θ = 91.4
sin2θ = 91.4*9.81/427^2 = .0049176
2θ = 0.2817° or 177.7183°
θ = 0.1409° or 88.859°
he drawing shows an exaggerated view of a rifl e that has been
“sighted in” for a 91.4-meter target. If the muzzle speed of the bullet is
υ0 = 427 m/s, what are the two possible angles θ1 and θ2 between the rifl e
barrel and the horizontal such that the bullet will hit the target? One of
these angles is so large that it is never used in target shooting. (Hint: The
following trigonometric identity may be useful: 2 sin θ cos θ = sin 2θ.)
1 answer
1 answer