You would like to shoot an orange out of a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 36.0 m/s at an angle of 30.0° above the horizontal from a height of 1.40 m while standing 41.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has travelled the 41.0 m horizontally to the tree?

horizontalvelocity=36*cos30

horizDistance=horveloicty*time
time= 41/36cos30

height attained:36sin30*timeabove-1/2 g timbabove^2

and adjust that for the starting height 1.4m.

u = 36 cos 30 = 31.18 m/s forever

d = u t = 41
so t = 41/31.18 = 1.315 seconds in air

Vi = 36 sin 30 = 18 m/s

h = Hi + Vi t - 4.9 t^2
h = 1.4 + 18(1.315) - 4.9 (1.315)^2
= 16.6 meters high

To determine the height of the arrow once it has traveled horizontally to the tree, we need to apply projectile motion equations.

Step 1: Analyze the given information and identify the relevant quantities:
- Initial velocity (V₀) of the arrow, which is 36.0 m/s
- Launch angle (θ) of the arrow, which is 30.0° above the horizontal
- Horizontal distance (d) from the archer to the tree, which is 41.0 m
- Vertical distance (h₀) between the archer and the ground, which is 1.40 m
- Vertical distance (h) of the orange above the ground, which is 5.00 m

Step 2: Break down the initial velocity into horizontal and vertical components:
The horizontal component of the initial velocity (V₀x) is given by:
V₀x = V₀ * cos(θ)
V₀x = 36.0 m/s * cos(30.0°)
V₀x = 31.18 m/s

The vertical component of the initial velocity (V₀y) is given by:
V₀y = V₀ * sin(θ)
V₀y = 36.0 m/s * sin(30.0°)
V₀y = 18.00 m/s

Step 3: Determine the time it takes for the arrow to reach the tree:
Since we can neglect air resistance, the time of flight (T) for the arrow is determined solely by the vertical motion and is given by:
T = (2 * V₀y) / g
T = (2 * 18.00 m/s) / 9.8 m/s²
T ≈ 3.67 s

Step 4: Calculate the vertical height (h₁) of the arrow at the tree:
To find the vertical height, we need to use the equation:
h = h₀ + V₀yt - (1/2) * g * t²

Plugging in the known values:
h = 1.40 m + (18.00 m/s)(3.67 s) - (1/2)(9.8 m/s²)(3.67 s)²
h ≈ 53.52 m

Therefore, the height of the arrow once it has traveled the 41.0 m horizontally to the tree is approximately 53.52 meters.