91Ó°ÊÓ

i) Friction is zero

ii) Height $h_1=5.0\text{cm}\&\text{ }{h}_2=1.60\text{cm}$and

iii) Horizontal distance,$d=6.00\text{cm}$

As the problem says there is no friction, potential and kinetic energy is conserved. Therefore, first find the change in kinetic energy of the ball during upward motion. In addition, by using initial height, define the time taken by the ball to reach that height. With the help of that time,find initial velocity at that point. In addition, by using all calculated parameters, find initial velocity of ball.

Formulae:

$K.E.=\frac{1}{2}M{v}^2$

$s=v_{0}t+\frac{1}{2}g{t}^2$

At initial position, the energy is conserved.

$\text{Initial}PE=0,$

$\text{Initial}KE=\frac{1}{2}×M×v_0^2$

As the ball moves to thetop, kinetic energy decreases and potential energy increases.

So,

$P.E.=M×g×h$

And new kinetic energy

$K.E.=\frac{1}{2}×M×v_0^2−M×g×h_1$

New kinetic energy at top

$K.E.=\frac{1}{2}×M×v_0^2−M×g×h_1$

We have,$h_1=5.00\text{cm}$and$g=9.80{\text{m/s}}^{\text{2}}$

$K.E.=\frac{1}{2}×M×v_0^2−M×0.049{\text{″¾}}^{\text{2}}{\text{/s}}^{\text{2}}$

As the ball leaves the plateau, its horizontal velocity remains constant. However, its vertical velocity increases from zero at the rate of$9.8\text{m/s}$per second. Now, we can findthetime to travel $0.016\text{″¾}$distance.

We can first find the time taken by the ball to travel at a height$h_2$from the top to the bottom. We can usethe second kinematic equation

$\begin{array}{c}{h}_2=v_0\text{t}+\frac{1}{2}g{t}^2\\ 0.016\text{″¾}=0+\frac{1}{2}×9.8{\text{″¾/s}}^{\text{2}}×t^2\end{array}$

$\begin{array}{c}t=\sqrt{\frac{0.016\text{″¾}}{4.9{\text{″¾/s}}^{\text{2}}}}\\ =0.058\text{sec}\end{array}$

Now we can determine the horizontal velocity to move$0.06\text{m}$intheabove time.

We apply second kinematic equation here again

$s=v_{0}t+\frac{1}{2}g{t}^2$

$0.06\text{″¾}=\left(v_0\right)0.058\text{ s}+0$

$v_0=\frac{0.06\text{″¾}}{0.058\text{s}}$

$⇒v_0=1.305\text{m}/\text{s}$

Therefore, we can find kinetic energy at height $h_2$as,

$K.E.=\frac{1}{2}×M×{\left(1.305\text{″¾/s}\right)}^2$

Since there is no friction, the K.E. ofthe ball at the top is equal to kinetic energy as it leaves the plateau,

$\begin{array}{c}\frac{1}{2}×M×v_0^2−M×0.049{\text{″¾}}^{\text{2}}{\text{/s}}^{\text{2}}=\frac{1}{2}×M×{\left(1.305\text{″¾/s}\right)}^2\\ \frac{1}{2}×v_0^2−0.049{\text{″¾}}^{\text{2}}{\text{/s}}^{\text{2}}=\frac{1}{2}×{\left(1.305\text{″¾/s}\right)}^2\\ v_0^2−0.098={\left(1.305\text{″¾/s}\right)}^2\\ v_0=1.34\text{m}/\text{s}\end{array}$