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anonymous
 4 years ago
A rocketship has an acceleration of 25 m/s^2 for 1 minute and is shot at 85 degrees from thehorizontal. How far from the launchpad will the rocket land. Neglect air resistance.
anonymous
 4 years ago
A rocketship has an acceleration of 25 m/s^2 for 1 minute and is shot at 85 degrees from thehorizontal. How far from the launchpad will the rocket land. Neglect air resistance.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Waite a minute please. brb for you.

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1i think we need to find velocity at the end of the acceleration and then just do the kinematics problem. split it into parts, the constant acceleration problem and then the kinematics problem. so acceleration is 25cos85 ihat and 25sin85 jhat (v=vo+at) so vx=0+25cos85*60=130.73 ihat vy=0+25sin85*60=1494.29 jhat

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1xx0=vo*t+1/2 at^2 so at the end of the acceleration we are 65.37m in the x direction and 747.15m in the air

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1we no longer have a x direction acceleration only a y direction acceleration, so now we figure out how long and how far it takes to get vy=0 from 1494.29 m/s with an acceleration of 9.8. then you add that distance to your height of 747.15m. Then calculate the time it would take to drop from that heighth. take the( time it takes to drop)+(the time from when the acceleration stopped to get to vy=0) then take that time, (xxo=130.73*(the T you added up) +(this is zero so it don't matter) take the x from this equation and add your 65.37m from the begining and you have your answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352956417082:dw First part of journey (with the given acceleration): on the yaxis, \(v_y=u_y+at\), where \(u_y=0\) \(v_y=(25sin 85^o9.8) (60)=906ms^{1}\) \(S_y=u_y t +\frac{1}{2}a_yt^2\) \(S_y=\frac{1}{2}(25sin 85^o9.8) (60)^2=27180m\) second part (\(a_y=9.8\)) \(v^2_y=u^2_y+2a_yS_y\), sub \(u_y=906, a_y =9.8, v_y=0\) \(S_y=41879.388m\) \(S_y=v_y t \frac{1}{2}a_yt^2\) \(41879.388=\frac{1}{2}(9.8) (t)^2\) \(t=92s\) For third part(\(u_y=0, a_y=9.8,S_y=27180+41879.388=69059.388\) \(S_y=u_y t +\frac{1}{2}a_yt^2\), Sub it all in again to find t. t=118s On the xaxis, for the first part,(\(u_x=0, a=25cos85^o, t=60\)) \(v=u_x +a_x t\) sub it all in to find \(v_x \) \(v_x = 2.179ms^{1}\) \(S_x=u_x t + \frac{1}{2}a_x t^2\). sub it all in again to find \(S_x\) \(S_x=65.3668m\) Then for the second and third part(\(u_x=2.179, a_x=0,t=92+118) \(S_x=u_x t + \frac{1}{2}a_x t^2\) Sub it all in to find \(S_x\) \(S_x=457.59\) \(S_{total}=(2.19+457.59)m\) Is there a part you need clarification?

Jusaquikie
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1352957134470:dw
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