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.