Work Done by the Gravitational Force — Question 4

Question 4

Landing speed can be independent of the launch direction

Take a uniform gravitational field with $g=10$ m/s$^2$, upward y positive, and neglect air resistance. Two projectiles leave the same point 5 m above the ground with equal speed 10 m/s but different launch directions. Neither hits any obstacle before landing. Only gravity acts.

1. Find gravity work per unit mass by landing.

2. Find the common landing speed.

3. Does the common landing speed imply equal flight times or equal landing velocities? Explain.

Question 4 - Solution

Landing speed can be independent of the launch direction

(a) Use the common height drop.

The governing relation is $$\boxed{\tfrac12(v_f^2-v_i^2)=g(y_i-y_f)}$$

$$W_g/m=g(y_i-y_f)=10(5)=50\ \mathrm{J/kg}$$ for each projectile.

(b) Apply the theorem per unit mass.

$$\tfrac12v_f^2-\tfrac12(10)^2=50,$$ giving $$v_f=\sqrt{200}=10\sqrt2\ \mathrm{m/s}.$$

(c) Retain the missing directional data.

No. Flight time depends on the initial vertical component, and the final horizontal component equals its own initial value. Equal speed magnitudes can accompany different velocity directions and different times.