Answer
given,
Mass of the cannon ball = 26 Kg
speed of cannon with muzzle = 962 m/s
angle with horizontal = 19.2°
angle of second ball = 144°
for first ball
[tex]v_y = v sin \theta_h[/tex]................(1)
[tex]v_y = (962) sin 19.2^0[/tex]
[tex]v_y = 316.37\ m/s[/tex]
now equating kinetic energy and potential energy
[tex]\dfrac{1}{2}mv_y^2 = mgh[/tex]
[tex]\dfrac{1}{2}v_y^2 = gh[/tex]
[tex]h = \dfrac{v_y^2}{2g}[/tex].................(2)
[tex]h = \dfrac{316.37^2}{2\times 9.8}[/tex]
h = 5106.63 m
total mechanical energy of ball at maximum height
angle = 144° = (90° + 54°)
so θ = 54°
[tex]E = \dfrac{1}{2}m(vsin\theta)^2[/tex]
[tex]E =\dfrac{1}{2}\times 26 \times (962\times sin 54^0)^2[/tex]
E = 7.75 x 10⁶ J