Recent content by kcalhoun

so use v^2=Vo^2+2aX?? But i still don't know what v or Vo are or how to get them...

i don't know how to find Vo with the given info.. that's my problem

vsin13?

yes 15m is the highest point i just don't know how to use that to solve the problem

Homework Statement The highest barrier that a projectile can clear is 15.0 m, when the projectile is launched at an angle of 13.0° above the horizontal. What is the projectile's launch speed? Homework Equations kinematics equations The Attempt at a Solution delta y=15m v0x=...

ohh nice yeah I am in arizona its only 830 here but i got 68.35degrees

ok i got part b figured out thanks a lot for youre help i appreciate it...

its velocity is 59.6m/s the x component is 59.6cos40 the y component is 59.6sin40

yeah i understand it that helped a lot thank you

so use the equation x=1/2(v0x+vx)t and i get 714.29m sweet so now part b...?

so it should be -7.827 and 15.645?

ok so after plugging the numbers into the quadratic equation i get 7.827s or -15.645s so t would bed 7.827s?

what i did was take the original equation y=v0yt+1/2ayt^2 and divided by t to get y=v0y+1/2ayt and the i solved for t to get t=(y-v0y)/(1/2ay) and then i plugged in the numbers.. t=(600-59.6sin40)/(1/2*9.8) so t=114.631 is that not right?

(600-59.6sin40)/(1/2*9.8)=114.631 does that look right?

right i used that equation to solve for t and i got 600=59.6sin40t+1/2(-9.8)t^2 which yields 114.631 right? then where do i go from there?