(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) (i) The time of flight for a projectile returning to the same horizontal level can be calculated using the formula: where is the initial velocity, is the angle of projection, and is the acceleration due to gravity.(ii) The horizontal range can be calculated using the formula: (iii) The equation of the trajectory for a projectile is given by: where is the vertical position, is the horizontal position, is the initial velocity, is the angle of projection, and is the acceleration due to gravity.(b) (i) The maximum height reached by a ball thrown at an angle with initial velocity can be calculated using the formula: where is the initial height of the roof.(ii) The time of flight for a ball thrown at an angle with initial velocity can be calculated using the formula: (a) (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.(ii) To show that the period of the satellite is given by , we can use the formula for the gravitational force between the satellite and the Earth: where is the gravitational constant, is the mass of the Earth, is the mass of the satellite, and 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: where is the orbital velocity of the satellite. Equating the two forces and solving for the period gives the desired result.(iii) If km, we can substitute this value into the formula for to find the period.