Projectile Question (1 Viewer)

trapizi · Oct 26, 2014

Most HSC past papers asked to find the maximum height of the projectile would reach from the equation:
$y=vtsin(\alpha )-\frac{1}{2}gt^2$
Show that
$y_{max}=\frac{v^2sin^2\alpha }{2g}$
My question is: Is it acceptable to use $\frac{\Delta _{y}}{-4a}$ to find maximum height?
Which is:
$\Delta _{y}=v^2sin^2\alpha$
$\therefore y_{MAX}=\frac{v^2sin^2\alpha }{{-4}\frac{-1}{2}g}=\frac{v^2sin^2\alpha }{2g}$

seventhroot · Oct 26, 2014

do you mean delta(y)/2a?

no you can't do that without a reason. you have to say "at max height, y dot = 0; therefore equate with zero" and do the whole subbing in

trapizi · Oct 26, 2014

seventhroot said:
do you mean delta(y)/2a?

no you can't do that without a reason. you have to say "at max height, y dot = 0; therefore equate with zero" and do the whole subbing in

Geez

seventhroot · Oct 26, 2014

trapizi said:
Geez

gg

Projectile Question (1 Viewer)

trapizi

╰(゜Д゜)╯

seventhroot

gg no re

trapizi

╰(゜Д゜)╯

seventhroot

gg no re

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