You would like to shoot an orange in a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. On your first try, you fire the arrow at 32.0 m/s at an angle of 30.0° above the horizontal from a height of 1.10 m while standing 52.0 m away. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has traveled the 52.0 m horizontally to the tree?
LOL thought you was slick!!! i don't get it either. Sapling is so annoying
To solve this problem, we can use the equations of motion for projectile motion. The vertical and horizontal coordinates of the arrow can be treated independently.
Let's first calculate the time it takes for the arrow to reach the tree. We'll consider the vertical motion equation:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
y = final vertical displacement (which is the height of the orange, 5.00 m)
y0 = initial vertical displacement (which is the height from where the arrow is fired, 1.10 m)
v0y = initial vertical component of velocity (v0 * sin(theta), where v0 is the initial velocity, and theta is the launch angle)
g = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values, we get:
5.00 m = 1.10 m + (32.0 m/s * sin(30.0°)) * t - (1/2)*(-9.8 m/s^2)*t^2
Simplifying the equation:
3.9 t^2 + 16 t - 3.90 = 0
Using the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 3.9, b = 16, and c = -3.9.
Plugging these values in the formula, we get:
t = (-16 ± √(16^2 - 4 * 3.9 * -3.9)) / (2 * 3.9)
Solving this equation yields two possible values for t: t1 = 1.138 s and t2 = -3.576 s. Since time cannot be negative in this context, the only meaningful solution is t1 = 1.138 s.
Now that we know how long it takes for the arrow to reach the tree, we can calculate the height of the arrow when it has traveled the 52.0 m horizontally. For the horizontal motion, we'll use the equation:
x = x0 + v0x * t
Where:
x = final horizontal displacement (which is 52.0 m)
x0 = initial horizontal displacement (which is 0 m)
v0x = initial horizontal component of velocity (v0 * cos(theta), where v0 is the initial velocity, and theta is the launch angle)
Plugging in the values, we get:
52.0 m = 0 m + (32.0 m/s * cos(30.0°)) * 1.138 s
Simplifying the equation:
52.0 m = 32.0 m/s * cos(30.0°) * 1.138 s
Dividing both sides by (32.0 m/s * cos(30.0°)):
t = 52.0 m / (32.0 m/s * cos(30.0°))
Calculating the right side of the equation:
t ≈ 1.791 s
Therefore, when the arrow has traveled 52.0 m horizontally to the tree, the height of the arrow would still be 1.10 m.
To find the height of the arrow once it has traveled the 52.0 m horizontally to the tree, we need to break down the projectile motion into horizontal and vertical components.
First, let's calculate the time it takes for the arrow to travel horizontally to the tree using its horizontal velocity.
Horizontal component:
The horizontal component of the initial velocity is given by Vx = V * cos(θ), where V is the initial velocity and θ is the angle of elevation.
Vx = 32.0 m/s * cos(30.0°)
Vx = 32.0 m/s * 0.866
Vx ≈ 27.712 m/s
Now, we can use the horizontal velocity and the distance traveled to calculate the time (t) it takes for the arrow to reach the tree.
Horizontal distance:
The horizontal distance (r) is given as 52.0 m.
Time:
Horizontal distance (r) = horizontal velocity (Vx) * time (t)
52.0 m = 27.712 m/s * t
Solving for t:
t = 52.0 m / 27.712 m/s
t ≈ 1.876 seconds
Now that we have the time it takes for the arrow to reach the tree, we can find the vertical height (y) at that time.
Vertical component:
The vertical component of the initial velocity is given by Vy = V * sin(θ).
Vy = 32.0 m/s * sin(30.0°)
Vy = 32.0 m/s * 0.5
Vy = 16.0 m/s
Using the time (t) and the vertical velocity (Vy), we can calculate the vertical height (y) of the arrow at the given time.
Vertical height:
Using the kinematic equation: y = V0y * t - (1/2) * g * t^2
where y is the vertical height, V0y is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time.
y = 16.0 m/s * 1.876 s - (1/2) * 9.8 m/s^2 * (1.876 s)^2
Simplifying the equation:
y = 30.016 m - (1/2) * 9.8 m/s^2 * 3.52 s^2
y = 30.016 m - 54.608 m
y ≈ -24.592 m
Since the height cannot be negative, the arrow would not be above the tree but below the ground level.
Therefore, the height of the arrow once it has traveled the 52.0 m horizontally to the tree would be approximately -24.592 meters.