Posted by Ian on Monday, October 3, 2011 at 10:53pm.
if the speed of the block is 9.5 m/s at the bottom of the ramp, from what height was the block released? what is the length of the ramp?" and i know that the mass is 7 kg, initial veloctity=0, it is frictionless, and inclined at an angle of 62 degreesif the speed of the black is 9.5 m/s at the bottom of the ramp, from what height was the block released? what is the length of the ramp?" and i know that the mass is 7 kg, initial veloctity=0, it is frictionless, and inclined at an angle of 62 degrees

Physics  Steve, Tuesday, October 4, 2011 at 5:36pm
the vertical acceleration is 9.8 * sin 62° = 8.652
s = 1/2 at² = 4.326 t²
v = at = 8.652t = 9.5, so
t = 1.098s
s = 4.326 * 1.098² = 5.22m
that's the height. The ramp length would be longer