A projectile is projected horizontally with a velocity u.Show that its trajectory is parabolic.Obtain expression for time of flight and horizontal range.

Sure, let's break this down:

1. Show that its trajectory is parabolic:

A projectile motion can be broken down into two components: horizontal motion and vertical motion.

• Horizontal motion: The horizontal velocity remains constant throughout the motion. This is because there is no horizontal force once the projectile is in flight. So, if we denote the horizontal velocity as u, the horizontal displacement (x) at any time (t) can be given as x = ut.

• Vertical motion: The vertical motion is influenced by the force of gravity. Therefore, it is a uniformly accelerated motion with acceleration equal to g (acceleration due to gravity). The vertical displacement (y) at any time (t) can be given as y = 1/2gt² (since the initial vertical velocity is zero in horizontal projection).

Now, if we eliminate t from these two equations, we get y = 1/2g * (x/u)². This is the equation of a parabola. Hence, the trajectory of the projectile is parabolic.

2. Obtain expression for time of flight and horizontal range:

• Time of flight (T): The time of flight is the time taken by the projectile to hit the ground. It is determined solely by the vertical motion. So, it is the time taken for an object to fall from the height h = 1/2gt², where h is the maximum height reached by the projectile. Solving this equation for t gives us T = sqrt(2h/g).

• Horizontal range (R): The horizontal range is the horizontal distance covered by the projectile. It is given by the product of the horizontal velocity and the time of flight. So, R = uT = u * sqrt(2h/g).

Please note that these derivations assume that the air resistance is negligible and the acceleration due to gravity is constant.