An arrow is shot from a bow at 20.0 m/s at an angle of 65 degrees above the horizontal. The arrow leaves the bow at a height of 1.80 m. At what height will it strike a wall that is 10.0 m away? (Remember to connect the two motions of the arrow using time.)

y = h + tanθ x - g/(2(vcosθ)^2) x^2

so, plug in x=10:

y = 1.8 + 2.144x - .580x^2
= 1.8+21.44-5.80
= 17.44

you can also use time, but it takes a bit more work:

Vx = 20 cosθ
use that to find t = 10/Vx
Vy = 20 sinθ
y = 1.8 + Vy t - 4.9t^2

If you churn it out, it should come out the same.

To solve this problem, we need to break down the motion of the arrow into horizontal and vertical components.

First, let's calculate the time it takes for the arrow to reach the wall. We can use the horizontal component of the velocity and the distance traveled to find the time.

1. Calculate the horizontal component of the velocity:
- The horizontal component of the velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
Where V is the initial velocity (20.0 m/s) and θ is the angle of the arrow above the horizontal (65 degrees).
- Substitute the values into the equation:
Vx = 20.0 m/s * cos(65°)
- Calculate Vx:
Vx ≈ 20.0 m/s * 0.42262 ≈ 8.4524 m/s

2. Calculate the time taken to reach the wall:
- The distance (d) can be divided by the horizontal velocity (Vx) to get the time (t):
t = d / Vx
Where d is the distance to the wall (10.0 m).
- Substitute the values into the equation:
t = 10.0 m / 8.4524 m/s
- Calculate t:
t ≈ 1.1833 s

Now that we have the time it takes for the arrow to reach the wall, we can calculate the height at which it will strike the wall.

3. Calculate the vertical height at t = 1.1833 s:
- We can use the vertical component of the velocity and the time to calculate the height.
- The vertical component of the velocity (Vy) can be found using the equation:
Vy = V * sin(θ)
Where V is the initial velocity (20.0 m/s) and θ is the angle of the arrow above the horizontal (65 degrees).
- Substitute the values into the equation:
Vy = 20.0 m/s * sin(65°)
- Calculate Vy:
Vy ≈ 20.0 m/s * 0.88536 ≈ 17.7072 m/s

- The height (h) can be calculated using the equation of motion:
h = Vy * t - 0.5 * g * t^2
Where g is the acceleration due to gravity (9.8 m/s^2).
- Substitute the values into the equation:
h = 17.7072 m/s * 1.1833 s - 0.5 * 9.8 m/s^2 * (1.1833 s)^2
- Calculate h:
h ≈ 20.948 m - 6.9203 m ≈ 14.028 m

Therefore, the arrow will strike the wall at a height of approximately 14.028 meters above the ground.

To solve this problem, we need to break down the motion of the arrow into its horizontal and vertical components.

First, let's find the time it will take for the arrow to reach the wall. We can use the horizontal component of its velocity for this. Let's denote the horizontal component as vx.

Given:
Initial velocity (v): 20.0 m/s
Angle of projection (θ): 65 degrees

The horizontal component (vx) is given by:
vx = v * cos(θ)

Calculating vx:
vx = 20.0 m/s * cos(65°)
vx ≈ 20.0 m/s * 0.4226
vx ≈ 8.452 m/s

Next, we can find the time of flight (t) using the horizontal distance (x) and horizontal velocity (vx) with the equation:
x = vx * t

Given:
Distance to the wall (x): 10.0 m

Calculating t:
10.0 m = 8.452 m/s * t
t ≈ 10.0 m / 8.452 m/s
t ≈ 1.184 s

Now that we have the time of flight, we can find the vertical component of the arrow's motion. Let's denote the vertical component as vy.

The vertical component (vy) is given by:
vy = v * sin(θ)

Calculating vy:
vy = 20.0 m/s * sin(65°)
vy ≈ 20.0 m/s * 0.9063
vy ≈ 18.126 m/s

Using the time of flight (t) and vertical component (vy), we can find the height (y) at which the arrow will strike the wall.

The height (y) is given by:
y = vy * t - 0.5 * g * t^2

Given:
Initial height (y₀): 1.80 m
Acceleration due to gravity (g): 9.8 m/s^2

Calculating y:
y = (18.126 m/s * 1.184 s) - (0.5 * 9.8 m/s^2 * (1.184 s)^2)
y ≈ 21.48 m - (0.5 * 9.8 m/s^2 * 1.401056 s^2)
y ≈ 21.48 m - 9.8 m/s^2 * 0.700528 s^2
y ≈ 21.48 m - 6.860616 m
y ≈ 14.619 m

Therefore, the arrow will strike the wall at a height of approximately 14.619 meters.