91Ó°ÊÓ

The mass of the puck is $m=1.0\mathrm{kg}$, the thrust of the puck is $F=2.0\mathrm{N}$, the distance traveled on friction less table is $s=4.0\mathrm{m}$, and the height of the table is $h=2.0\mathrm{m}$, the acceleration due to gravity is $g=9.8\mathrm{m}$.

We have to calculate the landing distance of the puck from the edge of the table.

The thrust of $2.0N$gives acceleration $a$to mass $m=1.0kg$ of puck:

$a=\frac{F}{m}$

$=\frac{2.0}{1.0}$

role="math" localid="1648367113682" $=2m/s^2$

Now puck is accelerated with $a$distance of $4.0m$.

The velocity of puck when it is at edge, using third equation of motion:

$v^2=u^2+2as$

$v$is the velocity at any time

$u$is the initial velocity$=0$

$a$is the acceleration

$s$is the distance travelled$=4.0m$

$v_f=\sqrt{2as}=\sqrt{2×2×4.0}$

role="math" localid="1648367610157" $v_f=4m/s$

At the edge, puck with horizontal motion starts vertical motion of falling downward with gravitational acceleration.

Using the second equation of motion:

$h=v_{i}t+\frac{1}{2}g{t}^2$

where,

$h=2.0m$

$g=9.8m/s^2$

$t^2=\frac{2h}{g}$

$t=\sqrt{\frac{2h}{g}}$

$t=\sqrt{\frac{2×2}{9.8}}$

$=0.63s$

As puck is accelerated horizontally still with$a=2m/s^2$, it goes distance$x$in the time$0.63s$, as it has horizontal velocity$v_i=4m/s$at the edge

Applying second equation of motion again:

$x=v_{i}t+\frac{1}{2}a{t}^2$

$x=(4.0×0.63)+\frac{1}{2}×2×{(0.63)}^2$

$x=2.91m$

So puck only goes $2.91m$away from base of the table

The puck only goes$2.91m$away from the edge of the table