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Maximum height of a projectile motion - Lab Report Example

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From the motion of the two balls projected horizontally, the balls are projected at the same initial velocity and same angle of elevation. The data was tabulated,…
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Maximum height of a projectile motion
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Since the gravitational pull acts downwards (vertical sense) then the horizontal and vertical components can be treated separately. Thus, the maximum height traveled by the body upwards can be calculated (Warren 73). For instance, a cannon shoots a ten kilograms cannon balls straight upwards, with a muzzle velocity of say 860 m/s, and then ignoring air resistance the maximum height attained by the cannon balls can be found. Taking the initial velocity in the vertical direction of the cannon ball to be vi and bearing in mind that the acceleration due to gravity is acting in the opposite direction (downwards).

At the point when the cannonball attains the maximum height, it has zero velocity thus vf = 0. At this point, the ball starts to accelerate downwards. The equation below is used to calculate the cannon’s highest point that it reaches before it starts falling downwards (Warren 72). Thus, the ball will go up to thirty-eight kilometers or approximately twenty-four miles (Serway, Jewett and Peroomian 20). Theoretically, the ball should come back and attain the velocity same as the launching velocity as it falls at the same point of launching, that is 860 m/s.

At each angle of elevation, the initial velocities were noted (McGinnis 73). The same procedure was repeated for the ball with a mass of 5.50 grams. The results were tabulated in a table as shown below: The result of the heights corresponding to every velocity was obtained, and curves were drawn for every mass. The results were as shown below. The table shows the results obtained for the ball of mass 9.5 g. The data obtained for the two balls was tabulated, and the maximum height reached by the ball at each launch relative to the angle of elevation was found (Breithaupt 26).

The maximum height was determined using the formulae in the equation (7). H max = …………………………………. (7). Each ball was launched at its time, and the muzzle velocities were maintained

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