MOSFET turn off time way too long, why?
As a check, remove C1, D1, and L1, and connect Q1.S directly to R2 so that Q1 drives a purely resistive load (R2). Now run your simulation and probe Q1's gate current.

You'll notice when voltage source V2 shuts off (transitions from +15V to 0V), the accumulated charge on Q1's gate discharges as current through R9, and Q1 remains ON until the gate charge level exits the Miller plateau, at which point the transistor goes into cutoff.

And of course, increasing R9's value slows down the discharge of Q1's gate, which causes Q1 to stay on longer after V2 transitions from +15V to 0V.

So what's the solution?  To expedite the discharge of Q1's gate and thereby turn off Q1 sooner, try this.  Replace voltage source V2 with a piece-wise linear voltage source (VPWL) and configure its time versus voltage values as follows:

time    voltage
 0s        0V
 50us      0V
 +5ns      15V
 50.2us    15V
 +10ns     -15V
 +40ns     -15V
 +5ns      0V


At t=50us, V2's voltage transitions from 0V to +15V, which turns on Q1. At t=50.2us, V2's voltage transitions from +15V to -15V, and V2's voltage remains at -15V for 40 ns, after which time V2's voltage rises from -15V back to zero volts.  This -15V 40ns pulse expedites the discharge of Q1's gate (reduces the Miller plateau transition time), and Q1 shuts off much sooner.

HINT: Research MOSFET &amp; IGBT gate driver ICs.