Muhammad must fire the arrow with a minimum speed of approximately 29.42 m/s to have it enter the window.
To solve this problem, we can use the principles of projectile motion. The arrow's motion can be divided into horizontal and vertical components. The horizontal component of the arrow's velocity remains constant, while the vertical component is affected by gravity. We aim to find the minimum initial velocity required for the arrow to enter the window at a height of 32.8 m.
Given:
Vertical displacement (window height) = 32.8 meters
Horizontal distance to the building = 63.6 meters
Launch angle = 51.5 degrees
Acceleration due to gravity (g) = 9.81 m/s²
We can break down the initial velocity (V₀) of the arrow into horizontal (V₀x) and vertical (V₀y) components:
V₀x = V₀ * cos(θ)
V₀y = V₀ * sin(θ)
Where θ is the launch angle in radians (θ = 51.5° * π / 180).
The time of flight (t) of the arrow can be determined from the vertical motion:
t = √(2 * Δy / g)
Where Δy is the vertical displacement (32.8 meters).
The horizontal distance covered by the arrow during this time can be found using the horizontal component of velocity:
Δx = V₀x * t
Since we want the arrow to hit the building at a horizontal distance of 63.6 meters, we need to set up an equation for Δx and solve for the initial velocity (V₀):
Δx = 63.6 meters
Δx = V₀x * t
63.6 = V₀ * cos(θ) * t
Substitute the expression for time (t) from the equation for the time of flight:
63.6 = V₀ * cos(θ) * √(2 * Δy / g)
Now, solve for V₀:
V₀ = 63.6 / (cos(θ) * √(2 * Δy / g))
Plug in the values:
θ = 51.5° * π / 180
Δy = 32.8 meters
g = 9.81 m/s²
Calculate V₀:
θ = 51.5° * π / 180 = 0.8988 radians
V₀ = 63.6 / (cos(0.8988) * √(2 * 32.8 / 9.81))
V₀ ≈ 29.42 m/s
So, Muhammad must fire the arrow with a minimum speed of approximately 29.42 m/s to have it enter the window.