Question

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:

Answer 1

U = 25J

Step-by-step explanation:

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:

`U=Mgh_(max)` (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:

`h_(max)=(v_o^2sin^2\theta)/(2g)` (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:

`U=Mg((v_o^2sin^2\theta)/(2g))=(1)/(2)Mv_o^2sin^2\theta\\\\U=(1)/(2)(2.0kg)(10m/s)^2(sin30\°)^2=25J`

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