An Arrow directed at a stationary target is shot at an angle of E25N. The arrow hits the intended target at the same height from which it is launched, 3.0s later.
Determine the distance from the archer to the target.
It matters how fast the arrow was launched, and at what elevation.
E25N ???
are you sure you do not mean 25 degrees above horizontal??
I will assume that is what you mean
S = initial speed
constant horizontal speed = u = S cos 25
initial up speed = Vi = S sin 25
distance d = u t = 3 S cos 25
time in air = t
hits max at t/2
0 = Vi - g (t/2) = Vi - 4.9 t in metric units
Vi = 4.9*3 = 14.7 meters/second
so
S sin 25 = 14.7
solve for S
u = S cos 25
solve for u
d = 3 u the end
To determine the distance from the archer to the target, we can break down the initial velocity of the arrow into its horizontal and vertical components and then use kinematic equations to calculate the distance.
Let's begin by breaking down the initial velocity of the arrow. The angle of E25N can be broken down into its horizontal and vertical components. "E25" means 25 degrees east of due north. This angle can also be expressed as 90 degrees - 25 degrees = 65 degrees north of due east. Therefore, the arrow is shot at an angle of 65 degrees north of due east.
Next, we can determine the horizontal and vertical components of the arrow's initial velocity. The horizontal component of velocity remains constant throughout the motion since gravity only affects the object vertically. The vertical component of velocity will be influenced by gravity and its value will change as the arrow's trajectory changes.
Let's denote:
V₀x as the horizontal component of the initial velocity,
V₀y as the vertical component of the initial velocity,
g as the acceleration due to gravity (approximately 9.8 m/s²).
Using trigonometry, we can find the values of V₀x and V₀y:
V₀x = V₀ * cosθ
V₀y = V₀ * sinθ
where V₀ represents the initial velocity of the arrow and θ represents the launch angle of 65 degrees north of due east.
Now, let's calculate the values of V₀x and V₀y. Since no values are given for the initial velocity V₀, we can assume it to be a known value.
Once we have the values of V₀x and V₀y, we can apply kinematic equations to determine the distance from the archer to the target. Since the vertical position of the arrow remains the same, we can use the equation:
V₀y = g * t
where t represents the time of flight, which is given as 3.0 seconds.
Rearranging the equation, we can solve for V₀y:
V₀y = g * t
Finally, using the equation for vertical displacement, we have:
Δy = V₀y * t + 0.5 * g * t²
Since the arrow hits the intended target at the same height from which it is launched, Δy will be equal to zero. Plugging in the values of V₀y, t, and g, we can solve for the distance from the archer to the target.