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 (yâ‚€): 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.