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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

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in that b and c
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Other answers:

check bottom page
i did a y=112 initial velocity
y=vit-4.9t^2
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is that right
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
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
instead of v=v0 cani use v=v0t+1/2gt^2
no, that equation is s = v0t + 1/2gt^2, and s is displacement, which we don't know yet

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