Question

A frictionless incline is 5.00 m long (the distance from the top of the incline to the bottom, measured along the incline). The vertical distance from the top of the incline to the bottom is 2.21 m. A small block is released from rest at the top of the incline and slides down the incline. (i) How long does it take the block to reach the ground?

Answer 1

Answer:

Time  taken by block to reach at lowest position is 0.67 sec.

Explanation:

Given that

Length,L=5 m

Height, h= 2.21 m

Given that surface is friction less and initially block is in rest.

As we know that acceleration due to gravity in vertical direction is g.

Lets take time taken by block to reach at lowest position is t.

So in vertical direction

$h=ut+\dfrac{1}{2}gt^2$

u= o ,initially at rest

$h=\dfrac{1}{2}gt^2$

$2.21=\dfrac{1}{2}\times 9.81\times t^2$

t= 0.67 s

Time  taken by block to reach at lowest position is 0.67 sec.