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You can shoot an arrow straight up so that it reaches the top of a 25-m-tall building. (a) How far will the arrow travel if you shoot it horizontally while pulling the bow in the same way? The arrow starts 1.45 m above the ground. (b) Where do you need to put a target that is 1.45 m above the ground in order to hit it if you aim 30° above the horizontal? (c) Determine the maximum distance that you can move the target and hit it with the arrow.

Respuesta :

Answer:

Part a)

[tex]d = 12 m[/tex]

Part b)

[tex]R = 43.27 m[/tex]

Part c)

[tex]R_{max} = 49.96 m[/tex]

Explanation:

As we can throw the arrow to maximum height of 25 m

so we will have

[tex]H = \frac{v^2}{2g}[/tex]

so we will have

[tex]25 = \frac{v^2}{2(9.81)}[/tex]

[tex]v = 22.14 m/s[/tex]

Part a)

If arrow is shot from 1.45 m height so the total time it took to reach the ground when it is shot horizontally given as

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

[tex]t = \sqrt{\frac{2(1.45}{9.81)}[/tex]

[tex]t = 0.54 s[/tex]

horizontal distance moved by arrow is given as

[tex]d = v t[/tex]

[tex]d = 0.54 \times 22.14[/tex]

[tex]d = 12 m[/tex]

Part b)

If arrow is shot at an angle 30 degree such that it will hit the target at same height

so the range of the arrow is given as

[tex]R = \frac{v^2 sin2\theta}{g}[/tex]

[tex]R = \frac{22.14^2 sin(2\times 30)}{9.81)[/tex]

[tex]R = 43.27 m[/tex]

Part c)

Now for maximum possible range the angle at which arrow is shot is given as 45 degree

so we will have

[tex]R_{max} = \frac{v^2}{g}[/tex]

[tex]R_{max} = \frac{22.14^2}{9.81}[/tex]

[tex]R_{max} = 49.96 m[/tex]

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