(a) A projectile is fired from the ground with an initial velocity u at an angle Theta to the horizontal. It returns to the same horizontal level.Find: (i) the time of flight. (5 Marks) (ii) the horizontal range , R. (4 Marks) (iii) the equation of the trajectory. (3 Marks) (b) A ball is thrown at 21m/s at 30^circ above the horizontal from the top of a roof 16 m high.Find: (i) the maximum height reached. (3 Marks) (ii) the time of flight. (5 Marks) Question 5 (a)(1) State Newton's law of universal gravitation. Mark) (ii) A satellite orbits the earth at a considerable distance R from the center of the earth in the plane of the equator and in the same direction of rotation as the earth. Show that the period of the satellite T is given by T^2=(4pi R^3)/(gr^2) where g is the gravitational acceleration and r is the radius of the earth. (8 Marks) (iii) If R=25000 km, find T. (2 Marks) 1

(a) <br />(i) The time of flight for a projectile returning to the same horizontal level can be calculated using the formula: <br />\[ T = \frac{2u \sin(\Theta)}{g} \]<br />where \( u \) is the initial velocity, \( \Theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.<br /><br />(ii) The horizontal range \( R \) can be calculated using the formula: <br />\[ R = \frac{u^2 \sin(2\Theta)}{g} \]<br /><br />(iii) The equation of the trajectory for a projectile is given by: <br />\[ y = x \tan(\Theta) - \frac{u^2 \sin^2(\Theta)}{2g} \]<br />where \( y \) is the vertical position, \( x \) is the horizontal position, \( u \) is the initial velocity, \( \Theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.<br /><br />(b) <br />(i) The maximum height reached by a ball thrown at an angle \( \Theta \) with initial velocity \( u \) can be calculated using the formula: <br />\[ H = \frac{u^2 \sin^2(\Theta)}{2g} + h \]<br />where \( h \) is the initial height of the roof.<br /><br />(ii) The time of flight for a ball thrown at an angle \( \Theta \) with initial velocity \( u \) can be calculated using the formula: <br />\[ T = \frac{2u \sin(\Theta)}{g} \]<br /><br />(a) <br />(i) Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.<br /><br />(ii) To show that the period of the satellite \( T \) is given by \( T^2 = \frac{4\pi R^3}{gr^2} \), we can use the formula for the gravitational force between the satellite and the Earth: <br />\[ F = \frac{GMm}{R^2} \]<br />where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( R \) is the distance between the satellite and the center of the Earth. The gravitational force provides the centripetal force required for the satellite to maintain its circular orbit: <br />\[ F = \frac{mv^2}{R} \]<br />where \( v \) is the orbital velocity of the satellite. Equating the two forces and solving for the period \( T \) gives the desired result.<br /><br />(iii) If \( R = 25000 \) km, we can substitute this value into the formula for \( T \) to find the period.