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Foolproof Method for Maths Questions
Sanaz:
Hey :) Not sure what I'm missing in this question. What is the working out for this question?
Cindy2k16:
--- Quote from: Sanaz on October 27, 2016, 03:38:15 pm ---Hey :) Not sure what I'm missing in this question. What is the working out for this question?
--- End quote ---
hi so you work out the vertical velocity after 2 seconds.
Since initial vertical velocity is 0 (it only has horizontal velocity initially) then using the formula v=u +at and subbing in u=0, a=9.8 and t=2, you get that the vertical velocity is 19.6m/s downwards.
Then you draw the triangle with the 45 angle and the 19.6 on the opposite side. Then using sintheta=O/H you sub in theta=45, O=19.6 and H is the resultant velocity. Rearrange the equation and you'll work out that the velocity is 27.7 m/s. So, D
Hope this helps
Sanaz:
--- Quote from: Cindy2k16 on October 27, 2016, 03:47:57 pm ---hi so you work out the vertical velocity after 2 seconds.
Since initial vertical velocity is 0 (it only has horizontal velocity initially) then using the formula v=u +at and subbing in u=0, a=9.8 and t=2, you get that the vertical velocity is 19.6m/s downwards.
Then you draw the triangle with the 45 angle and the 19.6 on the opposite side. Then using sintheta=O/H you sub in theta=45, O=19.6 and H is the resultant velocity. Rearrange the equation and you'll work out that the velocity is 27.7 m/s. So, D
Hope this helps
--- End quote ---
Oh yea haha now I realised where i went wrong, I was trying to find the x component of velocity and there wasn't enough info. But i didn't really need it :P Thanks for helping me!
strong1739:
how did you get tan?????? idk what im doing
RuiAce:
--- Quote from: strong1739 on October 29, 2016, 11:24:40 pm ---
how did you get tan?????? idk what im doing
--- End quote ---
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