Thanks I have a question regarding centripetal motion... I know acp=v2/R. So to to find it during ymax, I first started by finding the velocity it would be, which would be the horizontal component since the vertical component is 0, which got me 2.819 m/s (by solving vx=v0cosβ. But now I have no...
um no I don't have that equation but that's what I was sort of talking about in the second part of my reply
"I've tried that max height would occur at the time when vy=0, so I used the equation vy=v0sinβ-gt and set vy equal to zero to determine the time when the max was reached. For time I...
So I've been trying to calculate the y max height. I thought that would occur at .5t, which would give me .275 s. I then plugged this into the equation y-y0=(v0sinβ)t -.5gt2 so y-y0=-.088 and then added .9 which is y0. But this yields a number that is lower than the initial height, so makes no...
well i got the velocity x component by mulitplying initial velocity by cos(20) and got the y component by multiplying initial velocity by sin20 and then subtracting it by g(time). wait... do I take the sqrt of these added and squared? just remembered... so sqrt (vx2+vy2) right?
The time it would take the ball to travel, but I realized the error in calculating time. I now used the equation y0-y=v0(sinβ)t-.5(g)t2 and got .55 s for t. To get range, I then multiplied the time by vxt=3cos20(.55) and got 1.55 m. Sound about right? I am now trying to calculate the speed of...
Well to get the y max I first calculated time, which would be equal to -2v0/g, which got me .61 s. I then multiplied vxt which got me 1.72. How is that wrong?
Homework Statement
A baseball is bunted down the third base line leaving the bat from a height of .900m with initial v = 3.00 m/s at angle 20° above the horizontal.
I calculated that
x(range)= 1.72m
t(total)=.61s
vx=2.82 m/s
vy=1.03 m/s
-What is the speed of the ball before it hits the...