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Introductory Physics Homework Help
Kinematics: find maximum height
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[QUOTE="LifeMushroom, post: 6054466, member: 638602"] [h2]Homework Statement [/h2] A rocket takes off from the launch pad and moves directly upward with an acceleration of 29.4 m/s[SUP]2[/SUP]. It runs out of fuel after 4s and continues to coast upward, reaching a maximum height before falling back down to Earth. a) Find the rocket's maximum height. b) What is the velocity the instant before the rocket crashes back to the ground? [h2]Homework Equations[/h2] d = vi*t + (1/2)at[SUP]2[/SUP] vf[SUP]2[/SUP] = vi[SUP]2[/SUP] + 2ad vf = vi+at a = v/t [h2]The Attempt at a Solution[/h2] Maximum height is when the final velocity is zero, so: vf = 0 m/s, vi = 0m/s, a = 29.4 m/s^2 Using vf[SUP]2[/SUP] = vi[SUP]2[/SUP] + 2ad 0 = 0 + 2*29.4d 0 = 58.8d d = 0 a) I tried a=v/t and get a time of zero when rearranging it as well. This doesn't make any sense, how is the time zero and distance 0 at maximum height? I know this is likely meaning d and t = 0 at the beginning of the flight, but how can I find the max height? b) This sounds like instantaneous velocity or something, but I searched up the formula and it requires calculus, and our class isn't supposed to have calc in it. So I assume I have to use a kinematic equation, but I'm not sure which (either vf2 = vi2+2ad or vf=vi+at)... [/QUOTE]
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Kinematics: find maximum height
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