baileywohler
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You buy a plastic dart gun, and being a clever physics student
you decide to do a quick calculation to find its maximum
horizontal range.
You shoot the gun straight up, and it takes 4.0 seconds for the
dart to land back at the barrel (ignoring air resistance).
(a) What angle must your fire your gun to reach this max distance?
(b) What is the maximum horizontal range of your dart gun?
(c) What are the components of your initial velocity?

Respuesta :

C-consider a projectile launched with an initial velocity of 50 m/s at an angel of 60 degrees such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity.above the horizontal When we shoot the gun straight up, we can find the speed of dart when it leaves the gun: 0 = 2 Where – is a time of a dart’s flight. The horizontal range of an object shouted at angle: = 0 2 sin(2) And the maximum range at 45°: = 0 2 = ( 2 ) 2 = 2 4 = 9.81 2 ⋅ (5.6) 2 4 ≈ 77 Answer:: ≈ 77

Answer:

Part a)

[tex]\theta = 45 degree[/tex]

Part b)

[tex]R = 39.2[/tex]

Part c)

[tex]v_x = 13.86 m/s[/tex]

[tex]v_y = 13.86 m/s[/tex]

Explanation:

Since it took 4.0 s to come back at the same position so we can say

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]0 = v_y(4.0) - \frac{1}{2}(9.8)(4.0)^2[/tex]

[tex]v_y = 19.6 m/s[/tex]

Part a)

Now we know that horizontal range of projectile is given as

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

Now this range would be maximum if the angle of the projectile is giving maximum value of sine

so we have

[tex]sin(2\theta) = 1[/tex]

[tex]\theta = 45 degree[/tex]

Part b)

For maximum range we have

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

[tex]R = \frac{19.6^2}{9.8}[/tex]

[tex]R = 39.2[/tex]

Part c)

Since we projected at an angle of 45 degree

so the components are given as

[tex]v_x = 19.6 sin45 = 13.86 m/s[/tex]

[tex]v_y = 19.6 cos45 = 13.86 m/s[/tex]

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