An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path.
fired at angle T and speed s
u = s cos T
Vi = s sin T
range = 45 meters
Hi = 1.39
vertical problem:
v = Vi - 9.81 t
h = 1.39 + Vi t - 4.9 t^2
0 = 1.39 + Vi t - 4.9 t^2
horizontal problem:
u = s cos T forever
so
45 = u cos T * t
t = 45 / cos T
putting together:
0 = 1.39 + s sin T * t - 4.9 t^2
0 = 1.39 + s sin T * 45/cos T - 4.9 (45/cos T)^2
Now to continue you need another point. Like how high is A ???
I don't know, it's not given
To find the maximum height of the arrow along its parabolic path, we can use the fact that the path of a projectile, neglecting air resistance, follows a parabolic trajectory.
First, let's analyze the given information:
1. The arrow is released from a shoulder height of 1.39 m.
2. The arrow hits point A on the target, which is 18 m away.
3. When the target is removed, the arrow lands 45 m away.
We need to determine the maximum height of the arrow's path. Let's assume this maximum height as h_max.
To solve this problem, we can use the equations of motion for a projectile. These equations are:
1. Horizontal motion equation: x = v * t
2. Vertical motion equation: y = u * t + (1/2) * a * t^2
where:
- x and y are the horizontal and vertical distances traveled by the object, respectively.
- v is the horizontal velocity of the object.
- u is the initial vertical velocity of the object.
- a is the vertical acceleration, which for a projectile near the Earth's surface is -9.8 m/s^2 (assuming upward as positive).
Let's break down the problem into three parts:
Part 1: The arrow is launched and hits point A on the target.
Using the horizontal motion equation, we can find the time it takes for the arrow to travel 18 m:
18 = v * t1 Equation (1)
where v is the horizontal velocity of the arrow, and t1 is the time of flight.
Part 2: The target is removed, and the arrow lands 45 m away.
Using the horizontal motion equation again, we can find the time it takes for the arrow to travel 45 m:
45 = v * t2 Equation (2)
where t2 is the time of flight after the target is removed.
Part 3: Finding the maximum height.
To find the maximum height, we need to consider the vertical motion of the arrow. At the maximum height, the vertical component of the arrow's velocity becomes zero.
Using the vertical motion equation, we can determine the time it takes for the arrow to reach its maximum height:
0 = u + a * t_max Equation (3)
where t_max is the time it takes the arrow to reach its maximum height, and u is the initial vertical velocity of the arrow.
Finally, to find the maximum height, we can substitute the values of t_max, u, and a into the vertical motion equation:
h_max = u * t_max + (1/2) * a * t_max^2 Equation (4)
By solving Equations (1), (2), (3), and (4) simultaneously, we can determine the maximum height (h_max) of the arrow's path.
Note: This calculation assumes uniform acceleration due to gravity and neglects air resistance. In reality, air resistance and other factors may affect the arrow's trajectory.