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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2008 10:20:27 IST
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at what angle shoud a body be projected with a velocity 24m/s just to pass over an obstacle 16m high at a horizontal distance of 32m?
pls tell me da method.......
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i can feel the light betray me...............
not a single second left.......... |
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15 May 2008 10:22:43 IST
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i wud have replied u ..
but i know that seeing such an easy question ..a lot of ppl wud be posting answers ..
and u know wat ,..i am right
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SAIYANS ARE OF TRUE WARRIOR RACE . DONT UNDER-ESTIMATE US!
special theory of relativity ....i luv it...
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15 May 2008 10:23:25 IST
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u got d eqn of projectile...........
put dere x=32
.....y=16
fyn n u=24m/s
dis gives d ans .........
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nobody is wrong
even a stopped clock is right twice a day |
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15 May 2008 10:24:32 IST
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use
y = xtan@ - gx^2/2v^2cos@
16=32tan@ - g(32)^2/2v^2cos@
then use 1/cos^@ = sec^@ = 1+tan^@
u get a quadratic in tan@
solve for it
u get ur answer
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15 May 2008 10:29:19 IST
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take Range as twice of 32
and height as 16 m
and U as 24 m/s .
and solve for theta.
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---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly, neighborhood spideyunlimited |
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15 May 2008 11:14:59 IST
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use trajectory eq.
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I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
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