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a particle is projected from ground level with initial speed 8i+28j m/s. find the two times when its speed will be 10 m/s. i tried working this out by this.
√(8²+(21-gt)²)=100 (by the way the square root goes over it all even though it doesnt look like it) but when i solved for t i did not get the right answer. the right answer is 110/49 and 170/49.
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You made two mistakes in your formula: a) You used 21 instead of 28 and b) you used 100 instead of 10 (you can use 100 if you leave out the square root).
I assume that 28 is the vertical velocity (it looks like it in your approach).
Then we get: v(t) = 8i+(28-gt)
speed(t) = |v(t)| = sqrt(8²+(28-gt)²) = 10
8²+(28-gt)² = 100
(28-gt)² = 36
784-56gt+g²t² = 36
748-56gt+g²t² = 0
Just plug that into any formula for solving quadratics of your liking
If we set g as 9.8 you get exactly the answers given.
Last edited by Lehona (2016-10-23 10:13:50)
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Last edited by thickhead (2016-10-24 02:10:07)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Just a suggestion, but v and u are nearly impossible to distinguish in LaTeX, I'd usually avoid using them both at the same time.
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I hope it is clear now.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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