# What is projectile motion and its deivation explain

1
by pankajkumar

2014-10-28T03:27:56+05:30

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Let us say the the project is sent in to air at an angle of Ф with the horizontal and with an initial speed of u in that direction.

There is a deceleration in the vertical direction = -g.

The speed of projectile in the vertical direction is :  v = u + at =>
Vy = u sin Ф - g t        --  equation 1

The speed in the horizontal direction is :   Vx = u cos Ф.     --  equation 2
There is no acceleration in X-direction.

The distance travelled is :        s = ut+1/2 at²

=>    Sx = x = u (cos Ф) t         --- equation  3
t = x / (u cos Ф)

Sy = y = (u Sin Ф)  t  -  1/2 g t²

y =  u sinФ x / u cosФ - 1/2 g x² / u² Cos² Ф

y = x tan Ф - g x² / (2u²Cos²Ф)           ---   equation 4

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y = H when  Vy becomes 0.   =>  Vy = 0 = u sin Ф - g t
=>  t  =  u sin Ф / g
y = u Sin Ф * (u sin Ф / g)  - 1/2 g (u SinФ/g)²

y = H =  u² Sin² Ф / 2 g    --   equation  5

y = 0 =>  x = R

Substituting in equation 4, we get   R = u² Sin 2Ф / g       --  equation 6