You would like to shoot an orange out of a tree with your bow and arrow. The orange is hanging 5.00 m above the ground. You fire the arrow at 35.0 m/s oriented 30.0° above the horizontal from a height of 1.30 m while standing 47.0 m away on your first try. You may neglect air resistance. What is the height of the arrow above the ground once it has travelled the 47.0 m horizontally?
If you fire at the same speed and angle on your second try, how far away could you stand to hit the orange? (Hint: There are actually two correct solutions for this. One is if you hit the orange while the arrow is still on its way up, the other is if you hit it while the arrow is coming back down. You need to find both.)
the horizontal speed is 35 * cos30.0° = 30.3 m/s
So, it takes 47/30.3 = 1.55 seconds to travel.
So, plug that into the height
y = 1.3 + 35/2 t - 4.9t^2
to find the height at that time.
To find the correct distance, set
1.3 + 35/2 t - 4.9t^2 = 5.0
to find the time, then adjust the starting position.
Well, shooting an orange out of a tree seems like quite the fruity endeavor! Let's tackle this problem one juicy bit at a time.
First, let's find the time it takes for the arrow to travel the horizontal distance of 47.0 m. We can use the horizontal component of the arrow's initial velocity, which is v₀x = v₀ * cos(30.0°), where v₀ is the initial speed of 35.0 m/s. Plugging in the numbers, we get:
v₀x = 35.0 m/s * cos(30.0°)
Now, we can find the time it takes for the arrow to reach the orange by using the equation:
Δx = v₀x * t
Rearranging it, we have:
t = Δx / v₀x
Plugging in the values of Δx = 47.0 m and v₀x = 35.0 m/s * cos(30.0°), we can calculate the time. It might take a while, so feel free to peel a banana while waiting for the result!
Once you've got the time it takes for the arrow to reach the orange, you can use this time to find the height of the arrow above the ground. We know that the vertical component of the arrow's initial velocity is v₀y = v₀ * sin(30.0°), so the equation we can use is:
Δy = v₀y * t - 0.5 * g * t²
Where g is the acceleration due to gravity. But hey, since we're neglecting air resistance, we can all agree that oranges make terrible pilots, right?
Once you've determined the height of the arrow above the ground, congratulations! You've successfully found the answer to the first part of the question.
Now, onto the second part! If you fire at the same speed and angle on your second try, there are two possible distances from which you can hit the orange. One is when the arrow is on its way up, and the other is when the arrow is coming back down, much like a yo-yo.
To hit the orange on its way up, you can use the same approach as before, finding the time it takes for the arrow to reach the maximum height. Once you have that time, you can determine the horizontal distance by multiplying the time by the horizontal component of the arrow's initial velocity, v₀x.
To hit the orange on its way down, you can go with a similar approach. Find the time it takes for the arrow to reach the maximum height, then subtract that time from the total flight time. Multiply the remaining time by the horizontal component of the arrow's initial velocity, v₀x, and voila!
So, grab your bow, aim for the orange, and remember: even if you miss, you still have plenty of vitamin C-ing!
To solve this problem, we can first break down the initial velocity of the arrow into its horizontal and vertical components.
Given:
Initial vertical velocity (Vy) = 35.0 m/s * sin(30.0°)
Initial horizontal velocity (Vx) = 35.0 m/s * cos(30.0°)
Initial height (h) = 1.30 m
Distance to the tree (d) = 47.0 m
Step 1: Calculate the time of flight.
The time of flight can be found using the vertical motion equation:
h = Vy * t + 0.5 * g * t^2
Where g is the acceleration due to gravity, approximately 9.8 m/s^2. Since the orange's height remains constant throughout its flight, we neglect the vertical component gravity by setting it equal to zero.
0 = Vy * t + 0.5 * g * t^2
Substituting in the known values for Vy and g:
0 = (35.0 m/s * sin(30.0°)) * t + 0.5 * (9.8 m/s^2) * t^2
Rearranging the equation:
0 = 17.5 m/s * t + 4.9 m/s^2 * t^2
Step 2: Solve for the time of flight.
This equation is a quadratic equation in terms of t. By using the quadratic formula, we can find the time of flight.
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 4.9 m/s^2, b = 17.5 m/s, and c = 0.
Substituting the values into the quadratic formula:
t = (-(17.5 m/s) ± √((17.5 m/s)^2 - 4(4.9 m/s^2)(0))) / (2(4.9 m/s^2))
Simplifying the equation:
t = (-(17.5 m/s) ± √(306.25 m^2/s^2)) / (9.8 m/s^2)
Step 3: Calculate the time of flight.
We have two possible values for the time of flight, t1 and t2:
t1 = (-(17.5 m/s) + √(306.25 m^2/s^2)) / (9.8 m/s^2)
t2 = (-(17.5 m/s) - √(306.25 m^2/s^2)) / (9.8 m/s^2)
Calculating t1:
t1 = (-(17.5 m/s) + 17.5 s) / (9.8 m/s^2)
t1 = 0.0 s
Calculating t2:
t2 = (-(17.5 m/s) - 17.5 s) / (9.8 m/s^2)
t2 = -3.6 s
Since time cannot be negative, we discard t2.
Step 4: Calculate the final height of the arrow.
The final height (h_final) can be found using the equation:
h_final = h + Vy * t
Substituting the known values:
h_final = 1.30 m + (35.0 m/s * sin(30.0°)) * 0.0 s
Simplifying the equation:
h_final = 1.30 m
Therefore, the height of the arrow above the ground once it has traveled the 47.0 m horizontally is 1.30 m.
For the second part of the question:
Step 1: Calculate the maximum height reached by the arrow.
The maximum height can be found using the vertical motion equation:
Vy^2 = V0y^2 + 2gΔy
Where Δy is the maximum height reached.
Rearranging the equation:
Δy = (Vy^2 - V0y^2) / (2g)
Substituting in the known values:
Δy = (0 - (35.0 m/s * sin(30.0°))^2) / (2 * 9.8 m/s^2)
Simplifying the equation:
Δy = -147.06 m^2/s^2 / (19.6 m/s^2)
Δy = -7.5 m
Since height cannot be negative, we discard this negative value for the maximum height.
Step 2: Calculate the time taken to reach the maximum height.
The time taken to reach the maximum height can be found using the equation:
Vfy = V0y + gt
Where Vfy is the final vertical velocity (0 m/s).
Rearranging the equation:
t = (Vfy - V0y) / g
Substituting the known values:
t = (0 - (35.0 m/s * sin(30.0°))) / (9.8 m/s^2)
Simplifying the equation:
t = -17.5 m/s / (9.8 m/s^2)
t = -1.79 s
Since time cannot be negative, we discard this negative value for the time taken.
Step 3: Calculate the horizontal distance traveled.
The horizontal distance traveled can be found using the equation:
d = Vx * t
Substituting the known values:
d = (35.0 m/s * cos(30.0°)) * (-1.79 s)
Simplifying the equation:
d = -47.0 m
Therefore, you cannot stand at a distance of 47.0 m to hit the orange again.
To find the height of the arrow above the ground once it has traveled the 47.0 m horizontally, we can use the equations of projectile motion.
First, let's break down the initial velocity of the arrow into its horizontal and vertical components. The horizontal component can be found using the angle and speed:
Vx = V * cos(theta)
Vx = 35.0 m/s * cos(30.0°)
Vx = 35.0 m/s * 0.866
Vx = 30.11 m/s
The vertical component can be found using the angle and speed as well:
Vy = V * sin(theta)
Vy = 35.0 m/s * sin(30.0°)
Vy = 35.0 m/s * 0.5
Vy = 17.5 m/s
Next, we can use the time of flight equation to find the time it takes for the arrow to reach the horizontal distance of 47.0 m:
t = d / Vx
t = 47.0 m / 30.11 m/s
t ≈ 1.561 s
Now, we can use this time to find the height of the arrow above the ground. We need to find the vertical displacement at this particular time. We can use the equation for vertical displacement with constant acceleration:
y = yo + Vyo * t - (1/2) * g * t^2
Considering that the initial vertical velocity is positive (upward), and the gravitational acceleration is negative, we have:
y = 1.30 m + (17.5 m/s) * (1.561 s) - (1/2) * (-9.8 m/s^2) * (1.561 s)^2
y ≈ 14.031 m
Therefore, the arrow is at a height of approximately 14.031 m above the ground once it has traveled the 47.0 m horizontally.
Now let's move on to the second part of the question, determining how far away you could stand to hit the orange with the same speed and angle.
1. First, let's find the time it takes for the arrow to reach its maximum height (when it's still on its way up):
Vf = Vy + g * t_max
0 m/s = 17.5 m/s + (-9.8 m/s^2) * t_max
17.5 m/s = 9.8 m/s^2 * t_max
t_max ≈ 1.786 s
2. Using the time when the arrow reaches its maximum height, we can find the horizontal distance traveled:
d_max = Vx * t_max
d_max = 30.11 m/s * 1.786 s
d_max ≈ 53.803 m
So, you could stand approximately 53.803 m away to hit the orange while the arrow is still on its way up.
3. Next, let's find the time when the arrow returns to the same height as the orange:
y = yo + Vyo * t_return + (1/2) * g * t_return^2
0 m = 1.30 m + (17.5 m/s) * t_return + (1/2) * (-9.8 m/s^2) * t_return^2
This equation is quadratic in form, and we can solve it to find the time t_return. There will be two solutions since the arrow can be at the same height twice during its trajectory.
Using the quadratic formula: t_return = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = -(1/2) * g, b = 17.5 m/s, and c = 1.30 m:
t_return = (-(17.5 m/s) ± sqrt((17.5 m/s)^2 - 4 * (-(1/2) * 9.8 m/s^2) * 1.30 m)) / (2 * (-(1/2) * 9.8 m/s^2))
After solving the equation, we find two possible values for t_return:
t_return ≈ 3.068 s (coming back down)
t_return ≈ 0.388 s (still on the way up)
4. Finally, we can find the horizontal distance traveled for each of these times:
d_return = Vx * t_return
d_return1 = 30.11 m/s * 3.068 s
d_return1 ≈ 92.346 m (coming back down)
d_return2 = 30.11 m/s * 0.388 s
d_return2 ≈ 11.668 m (still on the way up)
Therefore, you could stand approximately 92.346 m away to hit the orange while the arrow is coming back down, or approximately 11.668 m away while the arrow is still on its way up.