(PART I) A Hollywood daredevil plans to jump the canyon shown in the figure on a
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Question
(PART I)A Hollywood daredevil plans to jump the canyon shown in the figure on a motorcycle. There is a 15. m drop and the horizontal distance originally planned was 60. m but it turns out the canyon is really 68.8 m across. If he desires a 3.2-second flight time, what is the correct angle for his launch ramp (deg)?
(above question just for reference)
(PART II)
What is the correct angle for his landing ramp (give a positive angle below the horizontal) and what is his predicted landing velocity? (Neglect air resistance.)
Explanation / Answer
range R = 68.8 m time of flight T = 3.2 s we know R = horizontal component of velocity * T from this horizontal component of velocity = R / T v cos = 21.5 -------( 1) In vertical direction : -------------------- Initial velocity u = v sin distance S = height = 15 m accleration a = -g = -9.8 m / s^ 2 time t = T from the relation S = ut + ( 1/ 2) at^ 2 15 = v sin *t - ( 1/ 2) gt^ 2 15 = 3.2 v sin - 50.17 3.2v sin = 65.17 v sin = 20.36 -----------( 2) eq( 2) / eq( 1) ==> v sin / v cos = 20.36 / 21.5 tan = 0.947 = 43.44 degrees plug it in eq( 1) we get v = 21.5/ cos 43.44 = 29.61 m / s (b). time taken to reach the maximum height t = v sin / g = 2.077 s So, time taken to reach the ground from maximum height t ' = T - t = 1.122 s vertical component of velocity when it landing V ' = gt ' = 11.0 m / s horizontal component of velocity when it landing V " = 21.5 m / s Since there is no accleration along horizontal direction So, land ing speed V = [ V'^ 2 + V"^ 2] = 24.15 m / s (c). landing angle = tan -1( V ' / V " ) = 27.09 degrees range R = 68.8 m time of flight T = 3.2 s we know R = horizontal component of velocity * T from this horizontal component of velocity = R / T v cos = 21.5 -------( 1) In vertical direction : -------------------- Initial velocity u = v sin distance S = height = 15 m accleration a = -g = -9.8 m / s^ 2 time t = T from the relation S = ut + ( 1/ 2) at^ 2 15 = v sin *t - ( 1/ 2) gt^ 2 15 = 3.2 v sin - 50.17 3.2v sin = 65.17 v sin = 20.36 -----------( 2) eq( 2) / eq( 1) ==> v sin / v cos = 20.36 / 21.5 tan = 0.947 = 43.44 degrees plug it in eq( 1) we get v = 21.5/ cos 43.44 = 29.61 m / s (b). time taken to reach the maximum height t = v sin / g = 2.077 s So, time taken to reach the ground from maximum height t ' = T - t = 1.122 s vertical component of velocity when it landing V ' = gt ' = 11.0 m / s horizontal component of velocity when it landing V " = 21.5 m / s Since there is no accleration along horizontal direction So, land ing speed V = [ V'^ 2 + V"^ 2] = 24.15 m / s (c). landing angle = tan -1( V ' / V " ) = 27.09 degreesRelated Questions
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