I shoot an arrow horizontally from a bow 1.25 m above the ground with an initial velocity of 46.1 m/s but I miss the target.

A) calculate how far away (horizontally from the release point) the arrow lands on the ground.

B) Calculate the magnitude and direction of the arrow's velocity just as it lands.

Correct A) D = 23.05 m  B) V = 46.37 m/s , α = 6.19°

Explanation:

The fired arrow performs a complex movement called a horizontal shot.

Horizontal shot is the result of the combination of two independent perpendicular movements, horizontal and vertical.

In the horizontal direction the arrow moves at a constant initial velocity

V₀ₓ and in the vertical direction performs a free fall without the initial velocity.

The free fall height is given by the formula

h = g · t²/2 , we take that the gravitational acceleration is g = 10 m/s²

From this formula we will calculate the time of fall, which is also the total time of movement of the arrow in the horizontal direction

t = √(2 h)/g = √(2 · 1.25)/10 = √0.25 = 0.5 seconds

The total distance the arrow reached

D = V₀ₓ · t = 46.1 · 0.5 = 23.05 m

The vertical component of the final velocity is

Vy = g · t = 10 · 0.5 = 5 m/s

We will calculate the total ultimate velocity using Pythagoras' theorem

V = √V₀ₓ² + Vy² = √46.1² + 5² = √2,125.21 + 25 = √2,150.21 = 46.37 m/s

The angle that the arrow will have with respect to the horizon or direction will be found using a tanα

tanα = Vy / V₀ₓ = 5 / 46.1 = 0.1084 => α = tan⁻¹ 0.1084 = 6.19°

α = 6.19°

God is with you!!!