An arrow is shot at an angle 32 degrees with the horizontal. It has a velocity of 46 m/s.?

Let’s determine the vertical and horizontal components of the velocity.

Vertical = 46 * sin 32, Horizontal = 46 * cos 32

As the arrow moves up to its maximum height, its vertical velocity decreases from 46 * sin 32 to 0 at the rate of 9.8 m/s each second. As the arrow falls from its maximum height, its vertical velocity increases 0 to 46 * sin 32 to 0 at the rate of 9.8 m/s each second. Since the distance up is equal to the distance down, the time up is equal to the time down. This means the total time is twice the time up. Let’s use the following equation to determine the time up.

vf = vi + a * t, vf = 0, vi = 46 * sin 32, a = -9.8

0 = 46 * sin 32 – 9.8 * t

t = 46 * sin 32 ÷ 9.8 = 2.487376138 seconds

Let’s use the following equation to determine the distance down.

d = vi * t + ½ * a * t^2, vi = 0, a = 9.8, t = 46 * sin 32 – 9.8

d = ½ * 9.8 * (46 * sin 32 ÷ 9.8)^2 = 30.31649626 meters

This is the maximum height of the arrow.

Let’s use the following equation to determine the horizontal distance.

Horizontal distance = Horizontal velocity * total time

Total time = 2 * 46 * sin 32 ÷ 9.8 = 92 * sin 32 ÷ 9.8 = 4.974752276 seconds

Horizontal distance = 46 * cos 32 * (92 * sin 32 ÷ 9.8) = 194.0661431 meters