Radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km h^{-1 }=\(900 \times \dfrac{5}{18}\)= 250 ms^{-1}

Centripetal acceleration: \(a_{c} = \dfrac{v^{2}}{r}\)

=\(\dfrac{\left (250 \right )^{2}}{1000}\) = 62.5 ms^{-2}

Acceleration due to gravity, g = 9.8 ms^{-2}

\(\dfrac{a_{c}}{g} = \dfrac{62.5}{9.8}\) = 6.38

\(a_{c}= 6.38 \; g\)

Answered by Pragya Singh | 1 year ago**(a) **Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

\(\theta (t)=tan^{-1}(\dfrac{v_{0y-gt}}{v_{ox}})\)

**(b)** Shows that the projection angle θ_{0} for a projectile launched from the origin is given by

\(\theta_{0}=tan^{-1}(\dfrac{4h_{m}}{R})\)

where the symbols have their usual meaning

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of a radius of 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?

A fighter plane flying horizontally at an altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^{-1} to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 m s^{-2} ).

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 be fixed, and neglect air resistance.

Can we associate a vector with

**(i)** a sphere

**(ii)** the length of a wire bent into a loop

**(iii)** a plane area

Clarify for the same.