91Ó°ÊÓ

The magnitude of an object's rate of change of location with time or per unit of time is its speed, it is thus a scalar number.

The motion of the puck is illustrated in the diagram below:

Here, $v_i$is the puck's starting velocity,$v_f$is the puck's final velocity, $d$is the ramp's length, and is $h$is the ramp's height in feet.

According to the diagram,

$\sin 20=\frac{h}{d}$

$h=d\sin 20$

Calculate the ramp's height as follows:

$h=d/\sin 20$

Putting$3.0\mathrm{m}$for $d$in $h=d/\sin 20$.

$h=(3.0\mathrm{m})/\sin 20$

$=1.026\mathrm{m}$.

The puck has merely kinetic energy at the bottom of the ramp.

At the bottom of the ramp, the potential energy is zero.

$P{E}_i=0$

At the bottom of the ramp, the kinetic energy is,

$K{E}_i=\frac{1}{2}m{v}_i^2$

When the ramp is at its highest point, $\left(h\right)$, because the puck only has potential energy, it has no kinetic energy. As a result, the puck's final speed is$0\mathrm{m}/\mathrm{s}\left(v_f=0\mathrm{m}/\mathrm{s}\right)$.

At the top of the ramp, the potential energy is,

$P{E}_f=mgh$

At the top of the ramp, the kinetic energy is zero,

$K{E}_f=0$

Apply the energy conservation law.

At the bottom of the ramp, the total energy is equal to the total energy at the top.

$\left(K{E}_i+P{E}_i\right)=\left(K{E}_f+P{E}_f\right)$

Putting $0$for $P{E}_i,\frac{1}{2}m{v}_i^2$for $K{E}_i,mgh$for $P{E}_f$, and $0$for $K{E}_f$in$\left(K{E}_i+P{E}_i\right)=\left(K{E}_f+P{E}_f\right)$

$\frac{1}{2}m{v}_i^2+0=0+mgh$

$\frac{1}{2}m{v}_i^2=mgh$

$v_i^2=2gh$

$v_i=\sqrt{2gh}$

Assume the puck's starting (minimum) speed is:

$v_i=\sqrt{2gh}$

Putting $1.026\mathrm{m}$for $h,9.8\mathrm{m}/{\mathrm{s}}^2$for $g$in $v_i=\sqrt{2gh}$.

$v_i=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(1.026\mathrm{m})}$\\

$=4.48437\mathrm{m}/\mathrm{s}$

The minimum speed of the puck is (rounded off to two major digits) $4.5\mathrm{m}/\mathrm{s}$.