A small block is projected up along an incline plane of inclination ${37}^o$ with velocity 24m/s. The co – efficient of friction of the block with the incline plane is 0.75. Assuming $g=10m/s^2.$ find the displacement (in meters) covered by the block is 3s 

The block will have its retardation while going up the inclined plane as:
$a=g\sin \theta +\mu g\cos \theta =10(\frac{3}{4}+\frac{0.75\times 4}{5})m/s^2=12m/s^2$ 
So, the block will stop after 2s. After the blocks stops, the frictional force will be sufficient to keep the 
block at rest as $\mu >\tan \theta .$ Thus, displacement only up to 2s will be same as displacement up to 3s.
$s=24m/s\left(2s\right)-\frac{1}{2}(12m/s^2){\left(2s\right)}^2=24m$