A 20.0-kg projectile is fired at an angle of 60⁰above the horizontal with a speed of 80.0 m/s. At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. Ignore air resistance. (a)How far from the point of firing does the other fragment strike if the terrain is level? How much energy is released during the explosion?

The given data is listed below as-

• The mass of the projectile is m = 20 kg
• The initial velocity of the projectile is V = 80.0 m/s
• Angle of projectile above the horizontal is$\theta =60°$

The kinetic energy of a particle equals the amount of work required to accelerate the particle from rest to speed V. Therefore, kinetic energy on the particle is given by-

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{V}}^{\mathbf{2}}$

The kinetic energy is a scalar and it is always positive or zero.

There are three stages during the motion of the projectile:

First stage- The motion of the projectile to its highest point

Second stage- the explosion

Third stage-motion of a fragment from explosion to the landing.

Now, take +x in the horizontal direction and +y as the upward movement.

The initial velocity is the vertical component of the velocity and the projectile moves with a negative acceleration for a vertical motion.

So, the equation of initial velocity is:

$$\begin{gathered} V_{iy}=80.0\sin 60° \\ =69.3m/s \end{gathered}$$

Now, the kinematic equation of displacement, is v = u + at.

$or,\hspace{0.33em}{v}_{fy}=v_{iy}+a_{y}t$

Substitute $V_{fY}=0,a_y=-9,8m/s^{-2}{V}_{iy}$ , in the above equation

$0=69.3-9.8\times t_1$

Therefore, $$\begin{gathered} t_1=\frac{69.3}{9.8} \\ {t}_1=7.07\hspace{0.33em}s \end{gathered}$$

The time taken by the projectile to reach its highest point is calculated as 7.07 s.

The horizontal component of initial velocity is equal to the horizontal motion of the projectile.

$$\begin{gathered} V_x=80\hspace{0.33em}\cos 60° \\ {V}_x=40\hspace{0.33em}m/s \end{gathered}$$

So,$x_1=v\cdot t_1$

$$\begin{gathered} x_1=40.0\times 7.07 \\ {x}_1=283\hspace{0.33em}m \end{gathered}$$

The second stage: The velocity of projectile before the explosion is:

$$\begin{gathered} \overrightarrow{v}=u_x\hat{i}+0\hat{j} \\ \overrightarrow{v}=40\hspace{0.33em}m/s\hat{i} \end{gathered}$$

Equation of momentum is

$$\begin{gathered} p=m\cdot v \\ p=(20.0)(40.0\hspace{0.33em}\hat{i})=(800\hspace{0.33em}kg\cdot m/s)\hat{i} \end{gathered}$$

The horizontal distance between land point to explosion point is

$$\begin{gathered} x_2=v_{Ax}·t_2 \\ =80\times 7.07 \\ =566m \end{gathered}$$

So, the horizontal distance between the firing point and the point where the other fragment land is:

$$\begin{gathered} d=x_1+x_2 \\ =566+283 \\ =849m \end{gathered}$$

Thus, the horizontal distance between the firing point and the point where the other fragment land is 849 m.

The energy released during explosion is the change in total kinetic energy of the system

$$\begin{gathered} ∆K=K_2-K_1 \\ =\left(K_A+K_B\right)-K_1 \\ =\frac{1}{2}{m}_A{V^2}_A+0-\frac{1}{2}m{V}^2 \\ ∆K=\frac{1}{2}\times 10\times {\left(80\right)}^2-\frac{1}{2}\times 20\times {\left(40\right)}^2 \\ =16000J \end{gathered}$$

Thus, the kinetic energy released during explosion is 16000 J.