Shooting a Steel Ball

Shooting a Steel Ball#

A small steel ball is shot at 5 \(m/s\) at a 48\(^{\circ}\) angle above the horizontal direction and its motion is well approximated by projectile motion.

Part 1#

When the ball returns to its original height, its velocity \(\overrightarrow{v} = (v_x, v_y)\) is:

Answer Section#

  • \((5\cos(48^{\circ}), \; -5\sin(48^{\circ}))\)

  • \((5\cos(48^{\circ}), \;5\sin(48^{\circ}))\)

  • \((5\sin(48^{\circ}), \; -5\cos(48^{\circ}))\)

  • \((5\sin(48^{\circ}), \;5\cos(48^{\circ}))\)

  • \((-5\cos(48^{\circ}), \; -5\sin(48^{\circ}))\)

  • \((-5\sin(48^{\circ}), \;5\cos(48^{\circ}))\)

Attribution#

Problem is licensed under the CC-BY-NC-SA 4.0 license.
The Creative Commons 4.0 license requiring attribution-BY, non-commercial-NC, and share-alike-SA license.