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Projectiles launched at an angle.

  • Thread starter Aracon504
  • Start date
  • #1
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Homework Statement



In 1991, Doug Danger rode a motorcycle to jump a horizontal distance of 76.5m. Find the maximm height of the jump if his angle with respect to the ground was 12.0?.

Homework Equations



Initial Velocity= ?g?x/2(sin ?)(cos ?)
?t= ?x/vi(cos ?)
?y= vi(sin ?)?t+1/2g(?t)^2

The Attempt at a Solution



vi= ?(9.81m/s/s)(76.5m)/2(sin 12?)(cos 12?)
vi= 43.3 m/s

?t= 76.5m/(43.3m/s)(cos 12?)
?t= 1.8s

?y= (43.3m/s)(sin 12?)(1.8s)+.5(-9.81m/s/s)(1.8s)^2



Not sure if this is right...
 
Last edited:

Answers and Replies

  • #2
I got V = 60.74 m/s and t = 2.575 s. Your answer of 43.3 m/s and 1.8 seconds will produce 76.23 meter horizontal distance.
I used Vcos(12)*t = 76.5 and 0 = Vsin(12) - gt to get t = Vsin(12)/g... so t = 0.0212V, multiplied by 2, t = 0.424V, then, V = 13.58 m/s, t = 5.758 s from here you may use

H = Vcos(12)t - 0.5gt^2
 
  • #3
8
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Thanks for the help.
 

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