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A body projected into the space and is no longer being propelled by fuel is called a projectile.
To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other:
(a) Along the vertical y-axis with a uniform downward acceleration 'g' and
(b) Along the horizontal x-axis with a uniform velocity forward.
Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:
Velocity along x-axis = ux = u cos θ
Acceleration along x-axis ax = 0
Velocity along y-axis = uy = u sin θ
Acceleration along y-axis ay = -g
Here we use different equation of motions of one dimension derived earlier to get the different parameters.
…... (a)
…... (b)
v2 = v02 – 2g (y-y0) …... (c)
When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b.
Therefore, 0 = (u sin θ) t - (½)gt2,
As t cannot equal to zero, then, total time of flight,
Horizontal Range (OA=X) = Horizontal velocity × Time of flight
= u cos θ × 2 u sin θ/g
So horizontal range,
At the highest point of the trajectory, vertical component of velocity is zero.
Therefore, 0 = (u sin θ)2 - 2g Hmax
So, maximum height would be,
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 θ.t and y = u sin θ.t - 1/2 gt2
Eliminating 't' form the above equations, we get,
y = x tan θ - (gx2/2u2cos2θ)
This is the equation of trajectory which is a parabola (y = ax + bx2).
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