An arrow is shot at an angle of θ=45∘
 above the horizontal. The arrow hits a tree a horizontal distance D=220m
 away, at the same height above the ground as it was shot. Use g=9.8m/s2
 for the magnitude of the acceleration due to gravity.Find  ta
 , the time that the arrow spends in the air.
Answer numerically in seconds, to two significant figures.

To find the time \( t_a \) that the arrow spends in the air, we can use the following kinematic equations. Since the arrow lands at the same height from which it was launched, we can focus on the horizontal motion.

Given:
- Initial angle \( \theta = 45^\circ \)
- Horizontal distance \( D = 220 \, \text{m} \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

We start by finding the horizontal component of the velocity. The horizontal range \( D \) can be calculated using the formula:

\[
D = v_0 \cdot t_a \cdot \cos(\theta)
\]

where \( v_0 \) is the initial velocity and \( t_a \) is the time of flight.

Next, we can express the time of flight using the vertical motion, where the arrow returns to the same height:

\[
t_a = \frac{2 v_0 \sin(\theta)}{g}
\]

Since \( \theta = 45^\circ \), \( \sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \), and we can express \( v_0 \) from the horizontal motion equation:

\[
v_0 = \frac{D}{t_a \cos(\theta)}
\]

Substituting this into the vertical motion equation:

\[
t_a = \frac{2 \cdot \frac{D}{t_a \cos(\theta)} \cdot \sin(\theta)}{g}
\]

Simplifying:

\[
t_a^2 = \frac{2 D \sin(\theta) \cos(\theta)}{g}
\]

Since \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \) and \( \theta = 45^\circ \), \( \sin(2\theta) = 1 \):

\[
t_a^2 = \frac{2D}{g}
\]

Finally:

\[
t_a = \sqrt{\frac{2D}{g}} = \sqrt{\frac{2 \times 220 \, \text{m}}{9.8 \, \text{m/s}^2}} 
\]

\[
t_a = \sqrt{\frac{440 \, \text{m}}{9.8 \, \text{m/s}^2}} \approx \sqrt{44.9 \, \text{s}^2} \approx 6.7 \, \text{s}
\]

So, the time the arrow spends in the air is approximately \( 6.7 \) seconds.

Would you like more details or have any questions? 

Here are 5 related questions to consider:
1. What would happen to the time of flight if the angle were increased to 60 degrees?
2. How would the range \( D \) change if the arrow were shot with a different initial speed?
3. What is the maximum height the arrow reaches during its flight?
4. How would air resistance affect the time of flight?
5. How would you calculate the velocity of the arrow when it hits the tree?

**Tip:** When working with projectile motion, always break the motion into horizontal and vertical components for easier calculation.

An arrow is shot at an angle of θ=45° above the horizontal. The arrow hits a tree a horizontal distance D=220m away, at the same height above the ground as it was shot. Use g=9.8m/s² for the magnitude of the acceleration due to gravity. Find ta, the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures.

Discover how to calculate the time an arrow spends in the air when shot at an angle of 45° with a range of 220 meters. This problem explores key concepts in projectile motion and kinematics, using trigonometric identities and kinematic equations. Ideal for students in Grades 11-12.

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