(a) the initial kinetic energy of the projectile is equal to:
[tex]K_i= \frac{1}{2}mv^2= \frac{1}{2}(10 kg)(500 m/s)^2=1.25 \cdot 10^6 J [/tex]
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
[tex]U_f=K_i= 1.25 \cdot 10^6 J [/tex]
b) If the projectile is fired with an angle of [tex]45^{\circ}[/tex], its velocity has 2 components, one in the x-direction and one in the y-direction:
[tex]v_x = v_0 \cos 45^{\circ} =(500 m/s) \cos 45^{\circ} =353.6 m/s[/tex]
[tex]v_y = v_0 \sin 45^{\circ} = (500 m/s)(\sin 45^{\circ} )=353.6 m/s[/tex]
This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
[tex]K_f = \frac{1}{2}m v_x^2 = \frac{1}{2} (10kg)(353.6 m/s)^2=6.25 \cdot 10^5 J [/tex]
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
[tex]K_i = K_f + U_f[/tex]
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:
[tex]U_f = K_i - K_f = 1.25 \cdot 10^6 J - 6.25 \cdot 10^5 J = 6.25 \cdot 10^5 J[/tex]
