Recent content by physicshelpme

i got 5.1 as the height

the angle is (.1) very small

so for the tan-1(2/20) of the little triangle be the same as theta of the whole triangle?

yea that makes more sense c watters

yeah your right the equations are not going to work. but if we are only considering the object in the y direction. I am unsure what to do next, this guy never teaches the math in class just talks about definitions something we can simply read at home. why is the service lime 20ft, and baseline...

thats exactly what he wrote i didn't change anything, b) so now we consider the height of the ball being struck v^2=vo^2 +2-g(y-yo) y=0 landing of the ball in the y direction?

0? if its linear then it has to be 0?

i did, how do i find a tracetory angle?

i think i got it now v^2=vo^2 +2-g(y-yo) vf=0m/s vi=? y=0 yo=2ft g=9.8?

how would i go about finding the trajectory with the horizontal?

ya that's what i wind up doing thanks man

i also did vf^2=vi^2 +2ay vi=0m/s a=-9.8m/s y=2ft->.61meters so vf=3.5m/s

yes, that's what i initially thought, i was thinking that at an initial velocity of 85km/h it changes 25m/s but that does not even make sense because how can velocity change a velocity wouldn't it have to be aceleration

so here what i think happened y=(vi x t)+1/2g x t^2 to find time vi=0m/s y=2ft g=-9.8m/s so we can find t but we need to convert to meters

picture