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Projectile motion

  1. Jul 18, 2006 #1
    ok, so i just had my exam 10 minutes ago and i cant think of anything else except this seemly easy problem that i couldnt get.

    a football is kicked on the horizontal plane (ie. y_0 = 0 ) at some angle alpha, it covers a horizontal displacement of 100 ft 2.5 seconds later, find the initial velocity and the angle. i feel so f-ing stupid, but i will greatly appreciate any help given
  2. jcsd
  3. Jul 18, 2006 #2


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    You know the horizontal velocity. Since the vertical velocity is independent, try figuring out high the football can go if it's to land in 2.5 seconds
  4. Jul 18, 2006 #3


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    Assume the initial speed is v. The initial horizontal speed is [itex]v cos(\alpha)[/itex], the initial vertical speed is [itex]v sin(\alpha)[/itex]. The (constant) vertical acceleration is -g and the there is 0 horizontal velocity so the horizontal speed is the constant [itex]vcos(\alpha)[/itex] and the vertical speed is [itex]-gt+ vsin(\alpha)[/itex]. Integrating those, the horizontal position is [itex]vcos(\alpha)t[/itex] and the vertical position is [itex]-(g/2)t^2+ vsin(\alpha)t[/itex]. Knowing that the horizontal distance covered in 2.5 sec. is 100 feet, gives [itex]2.5v cos(\alpha)= 100[/itex], You also know that the ball went up and back down to 0 in that time: the vertical equation gives [itex]-(g/2)(2.5)^2+ 2.5 vsin(\alpha)= 0[/itex]. That gives you two equations to solve for v and [itex]\alpha[/itex]
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