Recent content by aa2114

Would I have to find the time of flight for the whole range of the parabola first? time of flight: y = yo + 1/2gt^2 2.5 = 0 + 1/2(9.81)t^2 t = 0.7143 s Horizontal distance: x= xo + vit x = 0 + (12.03)(0.7143) x = 8.6 m

At max height v = 0 Time to reach max height: t = visin0/g t = 12.03sin50/9.81 t = 0.9394 seconds x= xo + vit x = 0 + (12.03)(0.9394) x = 11.3 m

Oops I think I got some of the signs wrong Acceleration: -mgsin0 - ukmgcos0 = ma a = -gsin50-(0.15)(9.81)cos50 a = -8.456 m/s^2 so velocity at the top: v^2 = vi^2 +2ad v = 14.14^2 + 2(-8.452)(3.264) v = 12.03 m/s

Homework Statement A spring with a spring constant = 1000 N/m is compressed 0.20 m then launches a block of mass 200 g. The horizontal surface is frictionless and the coefficient of kinetic friction with the inline is 0.15. What horizontal distance does the block cover while in the air after it...