91Ó°ÊÓ

The initial velocity of the projectile can be broken down into horizontal and vertical components. Using the angle of projection and initial speed:\[ v_{ix} = v_i \cos(\theta) = 80.0 \times \cos(60.0^\circ) = 40.0 \text{ m/s} \]The initial horizontal velocity \(v_{ix}\) is 40.0 m/s.

At the highest point, the vertical component of velocity is zero. Using the vertical motion equation:\[ v_{fy} = v_{iy} - gt \quad \Rightarrow \quad 0 = 80.0 \sin(60.0^\circ) - 9.8t \]Solving for \(t\):\[ t = \frac{80.0 \sin(60.0^\circ)}{9.8} = 7.07 \text{ s} \]

The total time of flight will be twice the time to reach the highest point:\[ T = 2 \times 7.07 = 14.14 \text{ s} \]

Since one fragment falls vertically with zero horizontal speed right after the explosion, it does not affect horizontal motion. Let's find the horizontal distance the second fragment travels from the point of explosion to ground.Given the explosion occurs at the highest point which splits the mass, second fragment's horizontal speed remains same, as there are no external horizontal forces:To find the distance traveled by the projectile till explosion:\[ x_{e} = v_{ix} \times 7.07 = 40.0 \times 7.07 = 282.8 \text{ m} \]After the explosion, the second fragment continues moving with the same horizontal velocity \(v_{ix}\) until it hits the ground:Total time from explosion to hitting ground (since launched from highest point, will fall back in same time):\[ 7.07s \]Distance after the explosion:\[ x_{e2} = v_{ix} \times 7.07 = 40.0 \times 7.07 = 282.8 \text{ m} \]The total horizontal range: \[ x_{total} = 282.8 + 282.8 = 565.6 \text{ m} \]

Initially, kinetic energy is conserved as only potential energy changes before explosion. At the highest point, potential energy increase equals decrease in initial total energy up there:\[ KE_i = \frac{1}{2}mv_i^2 \]Due to explosion, one fragment stops in vertical course (at max height):Half mass \(m = 10.0 \text{ kg}\):Kinetic Energy of stationary fragment \(KE_1=0\)Second fragment bombs down again (no height advantage):Energy before (Total Particle):New energy (as first is zero now) equals energy stored released due to one being nullified:\[ E_2 = \frac{1}{2} m v_f^2 + mgh\]Potential became true energy release due setting velocity zero:new speed partition also same due initial velocityThus, total potential more that realized pre explosion is energy released:\[ E_{release} = KE_i - E_2 \]

We've calculated the different components of the motion and accounted for the explosion scenario, allowing us to summarize the results.