An arrow directed at a stationary target is shot at an angle of [E 25 N]. The arrow hits the intended target at the same height from which it was launched, 3.0 s later.
a) Determine the distance from the archer to the target.
b) At what angle should the arrow be directed in to maximize its range?
The direction (25 degrees north of east) is useless information. You can ignore it. If it is in the air 3.0 seconds,it takes 1.5 seconds to rise to maximum height. The vertical velocity component is g*1.5s = 14.7 m/s
To compute how far the arrow goes, you need to know the launch angle A from horizontal OR the horizontal velocity component. Neither has been provided.
b) 45 degrees. The range is
(V^2/g)*sin(2A) That is a maximum when A = 45 degrees.
To solve part a), we can break down the initial velocity of the arrow into its horizontal and vertical components. The arrow is shot at an angle of 25° above the horizontal in the eastward direction (E 25 N).
Let's assume that the initial velocity of the arrow is "v". We can find the horizontal and vertical components of this velocity using trigonometry.
The horizontal component (v_x) can be found using the cosine function:
v_x = v * cos(angle)
The vertical component (v_y) can be found using the sine function:
v_y = v * sin(angle)
Since the arrow hits the target at the same height from which it was launched, the time it takes for the arrow to reach its peak height and then come down to the same height is half of the total time, which is 3.0 s.
Using the kinematic equation:
y = y_0 + v_y * t - 0.5 * g * t^2
where:
y = 0 (since the arrow returns to the same height)
y_0 = 0 (initial height)
g = acceleration due to gravity = 9.8 m/s^2
t = 3.0 s
We can solve for v_y using this equation:
0 = 0 + (v * sin(angle)) * (3.0) - 0.5 * (9.8) * (3.0)^2
Simplifying the equation and solving for v_y, we get:
v_y = 0.5 * (9.8) * (3.0)^2 / (3.0 * sin(angle))
Now, the total time of flight, T, can be found by doubling the time taken to reach the peak height:
T = 2 * 3.0 s = 6.0 s
The horizontal distance traveled by the arrow, D, can be calculated using the formula:
D = v_x * T
Substituting the values of v_x and T into the equation, we get:
D = (v * cos(angle)) * (6.0)
This is the distance from the archer to the target.
To solve part b), we need to find the angle that maximizes the horizontal distance traveled by the arrow. To do this, we can differentiate the equation for horizontal distance D with respect to the angle and then set the derivative equal to zero.
Differentiating the equation D = (v * cos(angle)) * (6.0) with respect to the angle, we get:
dD/d(angle) = -6.0 * v * sin(angle)
Setting the derivative equal to zero, we have:
-6.0 * v * sin(angle) = 0
Since v is a constant, we can solve for sin(angle) = 0, which gives us:
angle = 0 degrees
However, this angle does not make practical sense in this context. The other possible angle is 180 degrees, which means shooting the arrow in the opposite direction. The maximum range is achieved when shooting at an angle of 180 degrees, which is essentially shooting horizontally. The equation for horizontal distance becomes:
D = v * (6.0)
Therefore, to maximize the range of the arrow, the angle should be shot horizontally or at an angle of 180 degrees.