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along the vertical y-axis with a uniform downward acceleration ‘g’ and

• along the horizontal x-axis with a uniform velocity forward.

Consider a particle projected with an initial velocity u at an angle α with the horizontal x-axis as shown in figure 2.17. Velocity and accelerations can be resolved into two components:

Velocity along x-axis = u_x = u cos α

Acceleration along x-axis a_x = 0

Velocity along y-axis = u_y = u sin α

Acceleration along y-axis a_y = -g

Here we use different equation of motions of one dimension derived earlier to get the different parameters.

v² = v₀² – 2g (y – y₀)                                …(C)

Total Time of flight

When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation B.

Therefore,             0 = (u sin α) t – (½)gt²

or                   t = 2uSinα / g

(as t cannot equal to 0)

Horizontal Range

Horizontal Range (OA)  = Horizontal velocity × Time of flight

=        u cos α × 2u sin α/g

= u² sin 2α/g

=

Maximum Height

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore              0 = (u sin α)² – 2g H_max

or,           H_max = u² sin² α/ 2g

Equation of trajectory

Assuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the y-axis. Let P (x, y) be the position of the particle at instant after t second.

Then x = u cos α.t                and         y = u sin α.t – 1/2 gt²

Eliminating  ‘t’ form the above equations, we get,

y = x tan α – gx²/ 2u² cos² α

This is the equation of trajectory which is a parabola  (y = ax + bx²).

Illustrations

1.      A gun moving at a speed of 30m/sec fires at an angle 300 with a velocity 150m/s relative to the gun. Find the distance between the gun and the projectile when projectile hits the ground . (g = 10 m/sec)

Sol. Vertical component of velocity = 150 sin 300 = 75 m/sec

Horizontal component of velocity relative to gun = 150 cos 300

=                       =   75 √3 m/sec

Horizontal component of velocity relative to ground

=                              = 75 √3 + 30 ≈ 160 m/sec

Time of flight = (2 * 75 )/g = 15 sec

Range of projectile              = 160 × 15 = 2400 m

Distance moved by the gun and projectile = 2400 – 450 = 1950 m.

Consider a particle projected horizontally with a velocity u  from a point O as shown in figure 2.18.

Assuming the point of projection O as the origin of coordinates and horizontal direction as the X-axis and vertical direction as Y-axis. Let P (x, y) be the position of the particle after t seconds.

x = horizontal distance covered in time t = ut. …(1)

y = vertical distance covered in time t  = ½gt² ….(2)

Eliminate t from equations  (1) and (2) then

We get, y = (1/2 ) (g/u²) x²

This is the equation of parabola passing through the origin, with its vertex at the origin O. Hence the trajectory is a parabola.