91Ó°ÊÓ

1. The mass of the projectile is,$m=0.55\mathrm{kg}$.
2. The initial kinetic energy is,$K_1=1550$.
3. The maximum upward displacement is,$y=140\mathrm{m}$.
4. The vertical component of projectile velocity $v_y=65\frac{\mathrm{m}}{\mathrm{s}}$.

By using equation for kinetic energy Kand potential energy U and applying mechanical conservation of energy, we can find the horizontal and vertical components of the velocity of the projectile. For vertical displacement at a given vertical component of the velocity of the projectile we have to apply a kinematic equation to find its value.

Formula:

The kinetic energy is,

$K=\frac{1}{2}m{v_x}^2$

The potential energy is,$U=mgy$

The mechanical energy conservation,$K=K_1-U$

The kinematic equation is,${v_y}^2={v_{1y}}^2-2g∆y$

Write the expression for the kinetic energy as:

$$\begin{gathered} K=\frac{1}{2}m{v_x}^2 \\ ⇒K=\frac{1}{2}×0.55×{v_x}^2 \end{gathered}$$

Also, its potential energy is given by,

$$\begin{gathered} U=mgy \\ ⇒U=0.55×9.8×140 \\ ⇒U=755\mathrm{J} \end{gathered}$$

Thus, by mechanical energy conservation and solve:

$$\begin{gathered} K=K_1-U \\ =1550-755 \\ =795\mathrm{J} \end{gathered}$$

But, kinetic energy$K=\frac{1}{2}×0.55×{v_x}^2$ solve for the velocity as:

$$\begin{gathered} v_x=\sqrt{\frac{2K}{0.55}} \\ ⇒v_x=54\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

If$v_x=v_{1x}=54\frac{\mathrm{m}}{\mathrm{s}}$, so that the initial kinetic energy is solved as:

$$\begin{gathered} K_1=\frac{1}{2}m\left({v_{1x}}^2-{v_{1y}}^2\right) \\ ⇒{v_{1y}}^2=\left({54}^2-\frac{2×1550}{0.55}\right) \\ ⇒{v_{1y}}^2=52\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Applying kinematic equation,

$$\begin{gathered} {v_y}^2={v_{1y}}^2-2g∆y \\ ⇒∆y=\frac{{v_{1y}}^2-{y_y}^2}{2g} \\ ⇒∆y=\frac{{52}^2-{65}^2}{2×9.8} \\ ⇒∆y=-76\mathrm{m} \end{gathered}$$

The negative shows us it is below its launch point.