ksaimouli
  • ksaimouli
https://docs.google.com/viewer?a=v&q=cache:4ydUb6uwKLgJ:faculty.guhsd.net/teklund/Worksheet3-Projectile%2520Motion.pdf+ap+physics+projectile+motion+worksheet&hl=en&gl=us&pid=bl&srcid=ADGEESjxhBj5f-mP2GXXh28KT8TPcPsCtuUT3f4A0qhBvoPjnVrfr_wkFSFVDpa2CnysIct-AdRtuPE_xW2fgu2oBfyI2-J9SHGOu4ki-qX1YvOtnrjpq28525C_NRDhPdoUns0Ur-_g&sig=AHIEtbRGYXOpc3z7nsruZcqf9Iifnj-FzA
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ksaimouli
  • ksaimouli
in that b and c
ksaimouli
  • ksaimouli
@furnessj
anonymous
  • anonymous
do you have answers to check?

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ksaimouli
  • ksaimouli
check bottom page
ksaimouli
  • ksaimouli
i did a y=112 initial velocity
ksaimouli
  • ksaimouli
y=vit-4.9t^2
ksaimouli
  • ksaimouli
|dw:1352405626465:dw|
ksaimouli
  • ksaimouli
is that right
anonymous
  • anonymous
sorry i didn't look at the second page. First equation is simple: v = v_o +at 0 = 112 +(-g)t now, you need to change feet into metres to use 9.81 for g 112 / 3.28 = 34.146 so -34.146 / -9.81 = t
anonymous
  • anonymous
Second part: you are only worried about horizontal velocity (no air resistance), the up and down plays no part here. So simply: horizontal distance = horizontal speed x time hd = 224cos30 x 7
ksaimouli
  • ksaimouli
instead of v=v0 cani use v=v0t+1/2gt^2
anonymous
  • anonymous
no, that equation is s = v0t + 1/2gt^2, and s is displacement, which we don't know yet

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