Question

robin would like to shoot an orange in a tree with his bow and arrow. the orange is hanging ( y_1 = 5.00 , \text{m} ) above the ground. on his first try, robin releases the arrow at ( v_0 = 36.0 , \text{m/s} ) at an angle of ( \theta = 30.0^circ ) above the horizontal. the arrow has an initial height of ( y_0 = 1.50 , \text{m} ), and its tip is ( x = 54.0 , \text{m} ) away from the target orange. use ( g = 9.81 , \text{m/s}^2 ) for the acceleration due to gravity. treating the arrow as a point projectile and neglecting air resistance, what is the height ( h ) of the arrow once it has reached the horizontal position ( x ) of the orange?

Explanation:

Step1: Find horizontal component of velocity

The horizontal component of the initial velocity \( v_{0x} = v_0 \cos\theta \).
\( v_{0x} = 36.0 \cos(30.0^\circ) = 36.0 \times \frac{\sqrt{3}}{2} \approx 31.18 \, \text{m/s} \)

Step2: Calculate time to reach x

Using \( x = v_{0x} t \), solve for \( t \): \( t = \frac{x}{v_{0x}} \)
\( t = \frac{54.0}{31.18} \approx 1.732 \, \text{s} \)

Step3: Find vertical component of velocity

The vertical component of the initial velocity \( v_{0y} = v_0 \sin\theta \).
\( v_{0y} = 36.0 \sin(30.0^\circ) = 36.0 \times 0.5 = 18.0 \, \text{m/s} \)

Step4: Calculate vertical displacement

Using the equation \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \)
\( y = 1.50 + 18.0 \times 1.732 - \frac{1}{2} \times 9.81 \times (1.732)^2 \)
First, calculate \( 18.0 \times 1.732 = 31.176 \)
Then, \( \frac{1}{2} \times 9.81 \times (1.732)^2 \approx 0.5 \times 9.81 \times 3.0 \approx 14.715 \)
So, \( y = 1.50 + 31.176 - 14.715 \approx 17.961 \) Wait, no, wait: Wait, the orange is at \( y_1 = 5.00 \, \text{m} \)? Wait no, the question is to find the height \( h \) of the arrow when it reaches \( x = 54.0 \, \text{m} \). Wait, maybe I messed up. Wait, the initial height \( y_0 = 1.50 \, \text{m} \), and we need to find the height \( h \) (which is \( y \)) at \( x = 54.0 \, \text{m} \). Wait, but let's recalculate the time:

Wait, \( v_{0x} = 36 \cos 30 = 36 \times \frac{\sqrt{3}}{2} = 18\sqrt{3} \approx 31.176 \, \text{m/s} \)

Time \( t = \frac{54}{31.176} \approx 1.732 \, \text{s} \) (which is \( \sqrt{3} \approx 1.732 \), interesting)

Now, vertical displacement: \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \)

\( v_{0y} = 36 \sin 30 = 18 \, \text{m/s} \)

So, \( y = 1.50 + 18 \times 1.732 - 0.5 \times 9.81 \times (1.732)^2 \)

Calculate each term:

18 * 1.732 = 31.176

0.5 * 9.81 = 4.905

(1.732)^2 = 3.0 (since \( \sqrt{3}^2 = 3 \))

So, 4.905 * 3 = 14.715

Then, \( y = 1.50 + 31.176 - 14.715 = 1.50 + 16.461 = 17.961 \)? Wait, that can't be right, because the orange is at 5.00 m. Wait, no, maybe I misread the problem. Wait, the orange is at \( y_1 = 5.00 \, \text{m} \), but the question is to find the height of the arrow when it reaches the horizontal position of the orange (x=54.0 m). So maybe my calculation is wrong. Wait, let's check the time again. Wait, 36 cos 30 is about 31.18 m/s, so time to travel 54 m is 54 / 31.18 ≈ 1.732 s. Then vertical motion:

\( y = y_0 + v_{0y} t - 0.5 g t^2 \)

\( y_0 = 1.50 \, \text{m} \)

\( v_{0y} = 36 \sin 30 = 18 \, \text{m/s} \)

\( t = 54 / (36 \cos 30) = 54 / (18 \sqrt{3}) = 3 / \sqrt{3} = \sqrt{3} ≈ 1.732 \, \text{s} \)

So, \( y = 1.50 + 18 \times \sqrt{3} - 0.5 \times 9.81 \times (\sqrt{3})^2 \)

Calculate \( 18 \times \sqrt{3} ≈ 31.176 \)

\( 0.5 \times 9.81 \times 3 = 14.715 \)

So, \( y = 1.50 + 31.176 - 14.715 = 17.961 \, \text{m} \)? But the orange is at 5.00 m, so the arrow is way above? That seems possible? Wait, maybe the problem is to find the height relative to the ground, which is h = y. Wait, but let's check the calculation again.

Wait, maybe I made a mistake in the vertical velocity. Wait, 36 m/s at 30 degrees, so vertical component is 36 * sin(30) = 18 m/s, correct. Horizontal component 36 * cos(30) ≈ 31.18 m/s, correct. Time to travel 54 m: 54 / 31.18 ≈ 1.732 s, correct. Then vertical displacement:

\( y = y_0 + v_{0y} t - 0.5 g t^2 \)

\( y = 1.50 + 18*1.732 - 0.5*9.81*(1.732)^2 \)

Calculate 18*1.732: 18*1.732 = 31.176

0.5*9.81 = 4.905

(1.732)^2 = 3.0 (exactly, since 1.732 is approx sqrt(3), so (sqrt(3))^2 = 3)

So 4.905*3 = 14.715

Then 31.176 - 14.715 = 16.461

Then 1.50 + 16.461 = 17.961 m, which is approximately 18.0 m? Wait, but that seems high. Wait, maybe the problem is that the orange is at 5.00 m, but the arrow's height is 18 m, so it will pass over the orange. But the question is just to find the height h when it reaches x=54.0 m, regardless of the orange's height. So maybe that's correct.

Wait, let's do the calculation with more precise numbers.

\( v_{0x} = 36.0 \times \cos(30.0^\circ) = 36.0 \times \frac{\sqrt{3}}{2} = 18\sqrt{3} \approx 31.1763809 \, \text{m/s} \)

\( t = \frac{54.0}{31.1763809} \approx 1.7320508 \, \text{s} \) (which is exactly \( \sqrt{3} \) seconds, since \( 54 / (18\sqrt{3}) = 3 / \sqrt{3} = \sqrt{3} \))

\( v_{0y} = 36.0 \times \sin(30.0^\circ) = 18.0 \, \text{m/s} \)

Now, \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \)

\( y = 1.50 + 18.0 \times \sqrt{3} - 0.5 \times 9.81 \times (\sqrt{3})^2 \)

Calculate \( 18.0 \times \sqrt{3} \approx 31.1763809 \)

\( 0.5 \times 9.81 \times 3 = 14.715 \)

So, \( y = 1.50 + 31.1763809 - 14.715 = 1.50 + 16.4613809 = 17.9613809 \, \text{m} \)

So approximately 18.0 m? Wait, but let's check with g=9.81:

\( t = 54 / (36 \cos 30) = 54 / (36 * 0.8660254) = 54 / 31.1769144 = 1.732 \, \text{s} \)

Then, \( y = 1.50 + 18*1.732 - 0.5*9.81*(1.732)^2 \)

18*1.732 = 31.176

(1.732)^2 = 3.0

0.5*9.81*3 = 14.715

31.176 - 14.715 = 16.461

1.50 + 16.461 = 17.961, which is approximately 18.0 m. So the height h is approximately 18.0 m? Wait, but maybe I made a mistake in the problem interpretation. Wait, the orange is at y1=5.00 m, but the question is to find the height of the arrow when it reaches x=54.0 m, which is h = y (the height above ground). So the answer is approximately 18.0 m? Wait, but let's check again.

Wait, maybe the initial height is y0=1.50 m, and we need to find the height h, so h = y0 + vertical displacement. The vertical displacement is \( v_{0y} t - 0.5 g t^2 \). So:

Vertical displacement: \( 18*1.732 - 0.5*9.81*3 = 31.176 - 14.715 = 16.461 \)

Then h = 1.50 + 16.461 = 17.961 ≈ 18.0 m. So the answer is approximately 18.0 m.

Answer:

\boxed{18.0} (or more precisely, 17.96, but rounded to three significant figures, 18.0)