From the edge of a cliff, a 0.55 kgprojectile is launched with an initial kinetic energy of 1550 J. The projectile’s maximum upward displacement from the launch point is +140 m. What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is 65 m/s, what is its vertical displacement from the launch point?

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:

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

Also, its potential energy is given by,

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

Thus, by mechanical energy conservation and solve:

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

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

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

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

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

Applying kinematic equation,

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

The negative shows us it is below its launch point.