Posted by Tina on Wednesday, October 5, 2011 at 12:10am.
Vf^2 = Vo^2 + 2g*d,
Vo^2 = Vf^2 - 2g*d,
Vo^2 -= (23)^2 - (-19.6)*24 = 999.4
Vo = 3i.6m/s = Yo = ver. component of
initial velocity @ bottom of incline.
Vo = Yo / sinA = 31.6 / sin58=37.3m/s.
= Velocity at bottom of incline.
Related Questions
Physics - A mass of 18 kg is being pushed up a frictionless incline with a ...
Physics - A mass of 18 kg is being pushed up a frictionless incline with a ...
Physics 1 - A mass of 11 kg is being pushed up a frictionless incline with a ...
physics - [20 pts] A 2.00 kg block is pushed against a spring with negligible ...
PHYSICS PLEASE HELP !!!!! - Two objects are placed on a frictionless incline ...
Physics 1 - A mass of 5.2 kg lies at the top of a frictionless incline. It is ...
Physics - A mass of 5.6 kg lies at the top of a frictionless incline. It is ...
Physics - A mass of 5.6 kg lies at the top of a frictionless incline. It is ...
Physics 1 - A mass of 5.2 kg lies at the top of a frictionless incline. It is ...
physics - A mass of 5.6 kg lies at the top of a frictionless incline. It is ...
For Further Reading