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RSB [31]
4 years ago
5

A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling

at 98 m/s. If the acceleration of gravity is 9.81 m/s^2, what is the radius of curvature of the cannon balls path at this instant?
Engineering
1 answer:
ELEN [110]4 years ago
8 0

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity g=9.81m/sec^2

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by a_c=\frac{v^2}{r}, here v is velocity and r is radius of curvature

So \frac{98^2}{r}=9.81

r = 979 meter

So the radius of curvature is 979 meter

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A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i
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Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
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