Work Done by the Gravitational Force — Question 10

Question 10

Horizontal kinetic energy cannot fund extra ballistic height

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. A 2 kg projectile is launched with velocity $(12,5)$ m/s from y=0. Only gravity acts.

1. Find initial kinetic energy and kinetic energy at the apex.

2. Find gravity work to the apex and the maximum height.

3. Why does setting all the initial kinetic energy equal to $mgh$ give a wrong height?

Question 10 - Solution

Horizontal kinetic energy cannot fund extra ballistic height

(a) Keep the unchanged horizontal component.

The governing relation is $$\boxed{K=\tfrac12m(v_x^2+v_y^2),\qquad v_{x,f}=v_{x,i}}$$

$$K_i=\tfrac12(2)(12^2+5^2)=169\ \mathrm J.$$ At the apex $v_y=0$ but $$v_x=12\ \mathrm{m/s},$$ so $$K_{\mathrm{apex}}=144\ \mathrm J.$$

(b) Use only the actual kinetic-energy decrease.

$$W_g=144-169=-25\ \mathrm J.$$ Since $W_g=-20h$, the rise is $$h=1.25\ \mathrm m.$$

(c) Respect the component dynamics.

Gravity has no horizontal component, so it cannot remove the horizontal 144 J in free flight. The incorrect estimate $$169/20=8.45\ \mathrm m$$ assumes zero total speed at the apex, which this projectile never has.