A block is projected up a frictionless inclined plane with initial speed v0 = 3.

Question

A block is projected up a frictionless inclined plane with initial speed v0 = 3.48 m/s. The angle of incline is ? = 32.7°.
(a) How far up the plane does the block go?
answer = 1.14
(b) How long does it take to get there?
answer = ?
(c) What is its speed when it gets back to the bottom?
answer = 3.48 m/s
-for typing purposes, # = delta (change)
-So I just need help on B. I tried using the equation #x = Vi * t + 1/2 * a * t^2 and solving for t but it didn't work. For acceleration I used 5.39 because the force of gravity on an incline is g(sin(angle)) = 5.29. So yeah, thank you!

Explanation / Answer

the component of gravity acting down an inclined plane is g sin theta, so for this plane a=5.3m/s/s a) you can use vf^2=v0^2+2ad where vf=final speed =0 v0=initial speed=3.48m/s a=-5.3m/s solve for d: d=v0^2/2a=1.14m (this is the length up the ramp; 1.14 sin32.7=0.62m is the vertical displacement) b) d=v0t -1/2 at^2 where d=0.62m, a=-5.3m/s/s; solve quadratic for t c) if this is a frictionless plane, there is no less of energy, so the speed at the bottom is the same as the initial speed

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