FLIGHT OF PROJECTILE PDF

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FLIGHT OF PROJECTILE We shall assume the projectile to be moving without rotation in a vacuum. Such factors such as wind velocity, air resistance and rotation of the earth, which have an effect on the actual flight of the projectile, will be neglected. The principle of the solution is to resolve the...

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FLIGHT OF PROJECTILE We shall assume the projectile to be moving without rotation in a vacuum. Such factors such as wind velocity, air resistance and rotation of the earth, which have an effect on the actual flight of the projectile, will be neglected. The principle of the solution is to resolve the curvilinear motion of path into rectilinear motions along X and Y axis.

Instead of considering the actual path of the projectile, we combine its simultaneous projections upon in the X and Y axes. The equations of these rectilinear components of the path are found by substituting X and Y components of s, v, and a in the equations for rectilinear motion with constant acceleration as shown in table:

V = Vo + at s = Vot + 0.5at2 V2 = Vo2 + 2as

Other formulas: 1. Maximum height reached by the projectile

2. Range

3. Relationship of x and y

X component of flight ax = 0, Vox = Vocosθ

Y component of flight ay = -9.81, Voy = Vosinθ

Vx = Vox + axt Vx = Vocosθ x = Voxt + 0.5axt2 x = Vocosθ*t

Vy = Voy + ayt Vy = Vosinθ – gt y = Voyt + 0.5ayt2 y = Vosinθ*t - 0.5gt2 Vy2 = (Vosinθ)2 – 2gy

In the figure above, if the time of flight is less than that required to reach C, the projectile will be above its initial position and values of the Y displacement will be positive. If the time of flight is more than that required to reach C, the projectile will be on path CD and values of Y displacement will then be negative. At the topmost point of flight B, the value of Vy will be zero.

Examples: 1. A shell leaves a mortar with muzzle velocity of 150 m/s directed upward at 60 degrees with the horizontal. Determine position of shell and its resultant velocity 20 sec after firing. How high will it rise? 2. A cannon ball is fired from point A with horizontal muzzle velocity of 120 m/s. The cannon is located at elevation 60 m above ground, determine time for cannon to strike ball and range R. 3. A ball is thrown from a position 5 ft above ground to roof of a high rise building. If initial velocity of ball is 70 ft/s, inclined at angle of 60 degrees with horizontal, determine horizontal distance from point where ball is thrown to where it strikes the roof. 4. A stone is thrown from hill at angle 60 degrees with horizontal at initial velocity 30 m/s. After hitting level ground at base of hill, the stone covered a horizontal distance of 150 m. How high is the hill?

Source: Engineering Mechanics by Ferdinand Singer...