# An archer shoots an arrow with an initial velocity of 21 m/s straight up from his bow. He quickly reloads and shoots another arrow the same way 3 s later. At what time and height do the arrows meet?

The first arrow and the second meet at a height y.

The first arrow and the second meet at a height

The first one takes a time

We use the relationship of kinematics:

Where

So we have:

(I rounded a little bit....)

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To find the time and height when the arrows meet, you can use the kinematic equations. Since the first arrow is shot straight up, its motion is affected only by gravity. The second arrow's motion is the same once it's shot, but it starts 3 seconds later. Using the equation for displacement, (s = ut + \frac{1}{2}at^2), where (u) is initial velocity, (a) is acceleration due to gravity, and (t) is time, you can calculate the height and time of both arrows. The first arrow reaches its maximum height at (t = \frac{u}{a}). Substituting the values, you find (t_1 = \frac{21}{9.8} \approx 2.14) s, and its maximum height is (h_1 = ut_1 - \frac{1}{2}gt_1^2). For the second arrow, its time equation becomes (t_2 = t_1 + 3). The height it reaches is the same as the height the first arrow falls to after (3) seconds, so (h_2 = h_1 - \frac{1}{2}gt_2^2). Solving these equations gives (t_2 \approx 5.14) s and (h_2 \approx 64.1) m. Therefore, the arrows meet after approximately (5.14) seconds at a height of (64.1) meters.

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To find the time and height at which the two arrows meet, we need to analyze their vertical motion.

For the first arrow, we can use the kinematic equation:

[ s_1 = s_{0,1} + v_{0,1}t - \frac{1}{2}gt^2 ]

Where:

- ( s_1 ) is the displacement of the first arrow (which is the height)
- ( s_{0,1} ) is the initial displacement of the first arrow (which is 0 because it starts from the ground)
- ( v_{0,1} ) is the initial velocity of the first arrow (which is 21 m/s upwards)
- ( g ) is the acceleration due to gravity (which is approximately ( 9.81 , \text{m/s}^2 ))
- ( t ) is the time

For the second arrow, we use the same equation but with a different initial velocity and a time that is 3 seconds less:

[ s_2 = s_{0,2} + v_{0,2}(t-3) - \frac{1}{2}gt^2 ]

Where:

- ( s_2 ) is the displacement of the second arrow (which is the same as ( s_1 ) because both arrows meet at the same height)
- ( s_{0,2} ) is the initial displacement of the second arrow (which is 0 because it also starts from the ground)
- ( v_{0,2} ) is the initial velocity of the second arrow (which is 21 m/s upwards)
- ( g ) is the acceleration due to gravity (which is approximately ( 9.81 , \text{m/s}^2 ))
- ( t ) is the time

Since both arrows meet at the same height, we can equate ( s_1 ) and ( s_2 ) and solve for ( t ).

[ s_1 = s_2 ]

[ s_{0,1} + v_{0,1}t - \frac{1}{2}gt^2 = s_{0,2} + v_{0,2}(t-3) - \frac{1}{2}gt^2 ]

Solving this equation will give us the time at which the arrows meet. Once we have the time, we can substitute it back into either equation to find the height at which they meet.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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