Question

A 10-kg projectile is fired straight up with an initial velocity of 500 m/s. (a) What is the projectile’s potential energy at the top of its trajectory? (b) What would be the maximum potential energy if the projectile were fired at a 450 angle rather than straight up?

Answer 1

(a) the initial kinetic energy of the projectile is equal to:

`K_i= (1)/(2)mv^2= (1)/(2)(10 kg)(500 m/s)^2=1.25 \cdot 10^6 J`
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

`U_f=K_i= 1.25 \cdot 10^6 J`

b) If the projectile is fired with an angle of
`45^(\circ)`, its velocity has 2 components, one in the x-direction and one in the y-direction:

`v_x = v_0 \cos 45^(\circ) =(500 m/s) \cos 45^(\circ) =353.6 m/s`

`v_y = v_0 \sin 45^(\circ) = (500 m/s)(\sin 45^(\circ) )=353.6 m/s`

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:

`K_f = (1)/(2)m v_x^2 = (1)/(2) (10kg)(353.6 m/s)^2=6.25 \cdot 10^5 J`
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:

`K_i = K_f + U_f`
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:

`U_f = K_i - K_f = 1.25 \cdot 10^6 J - 6.25 \cdot 10^5 J = 6.25 \cdot 10^5 J`