A projectile of mass 2.0 kg is fired with an initial speed of 10 m/s at an angle of 30o above the horizontal. The potential energy of the projectile-Earth system when the projectile is at its highest point (relative to the potential energy when the projectile is at ground level) is:

In order to calculate the potential energy of the projectile-Earth system, when the projectile is at its maximum height, you use the following formula for the potential energy:

[tex]U=Mgh_{max}[/tex] (1)

M: mass of the projectile = 2.0kg

g: gravitational acceleration = 9.8m/s²

hmax: maximum height reached by the projectile.

The maximum height is given by the following formula:

[tex]h_{max}=\frac{v_o^2sin^2\theta}{2g}[/tex] (2)

vo: initial speed of the projectile = 10 m/s

θ: angle at which projectile was fires = 30°

You replace the expression (2) into the equation (1) and replace the values of all parameters:

[tex]U=Mg(\frac{v_o^2sin^2\theta}{2g})=\frac{1}{2}Mv_o^2sin^2\theta\\\\U=\frac{1}{2}(2.0kg)(10m/s)^2(sin30\°)^2=25J[/tex]

Hence, the potential energy of the projectile-Earth system is 25J