PREVNEXTSUBMITMenuResourcesGlossaryNotesClear and return to MenuSearch ResultsFilterSlide TextNotesSearch in:Check to includeSearch...TermsDefinitiondefinitionresourcessearchpauseplayreplaysubmitnextpreviousvolumeexample 2 a soccer player kicks a soccer ball so that the angle of incidence is 45° and the initial velocity is 23 m/s. find the following: maximum height total time of flight horizontal displacement velocity at impact click for audio the first step is to sketch out the problem to identify the known and unknown variables. a soccer player kicks a soccer ball so that the angle of incidence is 45° and the initial velocity is 23.0 m/s. find the following: maximum height (∆dymax) total time of flight (∆t) horizontal distance (∆dx) velocity at impact 45˚ 23.0 m/s ∆dx ∆dy ∆t 1. organize the variables into vertical and horizontal components 45˚ vix viy 23.0 m/s note: we can easily determine viy and vix using trigonometric ratios a) calculate the maximum height of the ball we don't have enough information initially to calculate the maximum height of the ball, but we can use the characteristics of parabolic motion to determine the variables we'll need. when the ball reaches its maximum height, the vertical velocity will be zero, so our first step is to determine the time it takes for the ball to reach its maximum height, setting vfy = 0. therefore, it takes 1.7 seconds to reach the maximum height. we can now use this value to determine the maximum height of the ball a) calculate the maximum height of the ball now that we know the time (∆t) it takes for the ball to reach its maximum height, we can solve for ∆dymax using the following equation: therefore, the maximum height (∆dymax) of the ball is 13.5 meters. b) total time of flight for this type of projectile motion problem, in which the launching points and landing points are the same we can employ some logic to determine the total time of flight. in this case, we can simply double the time it takes for the ball to reach its maximum height (apex) because it will take an equal amount of time to travel from the ground to the apex as it will from the apex to the ground. therefore the total time of flight ∆ttotal = 2∆t(at ∆dymax) ∆ttotal = 2(1.7 s) ∆ttotal = 3.4 s therefore the total time of flight is 3.4 seconds