anonymous
  • anonymous
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.
Physics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Waite a minute please. brb for you.
Jusaquikie
  • Jusaquikie
i 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 i-hat and 25sin85 j-hat (v=vo+at) so vx=0+25cos85*60=130.73 i-hat vy=0+25sin85*60=1494.29 j-hat
Jusaquikie
  • Jusaquikie
x-x0=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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Jusaquikie
  • Jusaquikie
we 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, (x-xo=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
  • anonymous
|dw:1352956417082:dw| First part of journey (with the given acceleration): on the y-axis, \(v_y=u_y+at\), where \(u_y=0\) \(v_y=(25sin 85^o-9.8) (60)=906ms^{-1}\) \(S_y=u_y t +\frac{1}{2}a_yt^2\) \(S_y=\frac{1}{2}(25sin 85^o-9.8) (60)^2=27180m\) 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 x-axis, 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
  • Jusaquikie
|dw:1352957134470:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.