Explanation Formula used: 1. Ideal Projectile Motion Equation : “The vector equation of the path for ideal projectile motion is r = ( υ 0 cos α ) t i + ( ( υ 0 sin α ) t − 1 2 g t 2 ) j , where υ 0 is the initial speed and α denotes the angle of elevation. Also the above equation gives the parametric equations x = ( υ 0 cos α ) t and y = ( υ 0 sin α ) t − 1 2 g t 2 , where x is the distance downrange and y is the height of the projectile at time t ≥ 0 .” 2. Height for Ideal Projectile Motion : “The maximum height is given by the equation y max = ( υ 0 sin α ) 2 2 g .” Calculation: It is known that, the time taken by the projectile to attain maximum height is t max = υ 0 sin α g . Compute the three-quarters of maximum height as shown below. y max = ( υ 0 sin α ) 2 2 g 3 4 y max = 3 4 ( υ 0 sin α ) 2 2 g = 3 ( υ 0 sin α ) 2 8 g Now substitute y = ( υ 0 sin α ) t − 1 2 g t 2 in the above equation as follows. ( υ 0 sin α ) t − 1 2 g t 2 = 3 ( υ 0 sin α ) 2 8 g 8 g ( ( υ 0 sin α ) t − 1 2 g t 2 ) = 3 ( υ 0 sin α ) 2 8 g ( υ 0 sin α ) t − 4 g 2 t 2 − 3 υ 0 2 sin 2 α = 0 4 g 2 t 2 − 8 g ( υ 0 sin α ) t + 3 υ 0 2 sin 2 α = 0 By quadratic formula, obtain the value of t as follows

4th Edition

ISBN: 9780134439099

Author: Hass, Joel., Heil, Christopher , WEIR, Maurice D.

Publisher: Pearson,

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Chapter 13.2, Problem 48E

To determine

To show: The time taken by the projectile to attain three-quarters of maximum height is half the time it takes to attain the maximum height.

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Chapter 13.2, Problem 47E

Chapter 13.3, Problem 1E

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