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A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?


Speed of the cyclist, v = 27 km/h = 7.5 m/s

Radius of the circular turn, r = 80 m

Centripetal acceleration is,

     
The question can be illustrated in the figure as shown below:


Consider, the cyclist begins cycling from point P and moves toward point Q. 

At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.

The angle between ac and aT is 900. Therefore, the resultant acceleration a is given by,


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(b) Shows that the projection angle θ0 for a projectile launched from the origin is given by,
straight theta subscript straight o space equals space tan to the power of negative 1 end exponent open parentheses space fraction numerator 4 straight h subscript straight m over denominator straight R end fraction close parentheses
where the symbols have their usual meaning.


Maximum space vertical space height. space straight h subscript straight m space equals fraction numerator space straight u subscript 0 squared space Sin squared straight theta space over denominator space 2 straight g end fraction space space space space space space space... space left parenthesis straight i right parenthesis

Horizontal space Range comma space straight R space equals space fraction numerator straight u subscript 0 to the power of 2 space end exponent Sin squared 2 straight theta space over denominator space straight g end fraction space space space space space space space space space space space space space... space left parenthesis ii right parenthesis

fraction numerator straight h subscript straight m space over denominator straight R end fraction equals space fraction numerator Sin squared straight theta space over denominator 2 Sin squared 2 straight theta end fraction space

space space space space space space space space equals space fraction numerator Sin space straight theta space cross times space Sin space straight theta space over denominator 2 space cross times 2 SinθCosθ end fraction

space space space space space space space space equals space fraction numerator sin space straight theta over denominator 4 space Cosθ end fraction space equals space tan space straight theta over 4

rightwards double arrow space tan space straight theta space equals space left parenthesis fraction numerator 4 hm over denominator straight R end fraction right parenthesis

rightwards double arrow space space straight theta space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 4 hm over denominator straight R end fraction close parentheses
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A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.

Given,

Range, 
R = 3 km

Angle of projection, θ = 30° 

Acceleration due to gravity, g = 9.8 m/s

Horizontal range for the projection velocity u0, is given by the relation: 

       R = u02 Sin 2θ / 

i.e.,  3 = u02 Sin 600 / g 

         u02 / g = 2√3                                    ...(i) 

The maximum range (Rmax) is achieved by the bullet when it is fired at an angle of 45° with the horizontal, that is, 

           Rmax = u02 / g                                 ...(ii)

On comparing equations (i) and (ii), we get:

Rmax = 3√

        = 2 X 1.732 = 3.46 km 

Hence, the bullet will not hit a target 5 km away.
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Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by,
                              


Let v0x and v0y respectively be the initial components of the velocity of the projectile motion along horizontal (x) and vertical (y) directions. 


Let vx and vy be the horizontal and vertical components of velocity at a point P respectively. 

Time taken by the projectile to reach point P = t

Using the first equation of motion along the vertical and horizontal directions, we get

vy = v0y = gt, and

vx = v0x


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