Explanation Given: The object is launched from the initial poition 〈 0 , y 0 〉 above horizontal ground with an initial velocity | v 0 | 〈 cos α , sin α 〉 . Definition used: “An object travels horizontally on ground and acts only by the gravitational force has an initial position 〈 x 0 , y 0 〉 = 〈 0 , 0 〉 and initial velocity 〈 u 0 , v 0 〉 = 〈 | v 0 | cos α , | v 0 | sin α 〉 . The trajectory, which is a segment of a parabola has properties as follows, Time of flight: T = 2 | v 0 | sin α g Range of the object: R = ( | v 0 | cos α ) T Maximum height object of the object: y ( T 2 ) = ( | v 0 | sin α ) 2 2 g ”. Calculation: The position vector of an object is 〈 v 0 cos α , v 0 sin α − g t 〉 . From the definition, initial velocity 〈 | v 0 | cos α , | v 0 | sin α 〉 is 〈 v 0 cos α , v 0 sin α − g t 〉 Compute the position vector of velocity 〈 v 0 cos α , v 0 sin α − g t 〉 . r ( t ) = ∫ v ( t ) d t = ∫ 〈 v 0 cos α , v 0 sin α − g t 〉 d t = 〈 v 0 t cos α , v 0 t sin α − g t 2 2 〉 Let us assume that the object hits the ground at ( a , 0). Campare the point ( a , 0) with the position vector 〈 v 0 t cos α , v 0 t sin α − g t 2 2 〉 . Then, v 0 T cos α = a (1) y 0 T sin α − g T 2 2 = 0 (2) Obtain the value of T from the equation (1), where T is the time of flight

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Author: Briggs

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Chapter 14.3, Problem 73E

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To derive: The formulas for time of flight, the range and maximum height.

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Chapter 14.3, Problem 72E

Chapter 14.3, Problem 74E

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