Projectile Motion — Question 2

Question 2

A horizontal launch and a drop land together

Two small balls start simultaneously 20 m above a level floor. Ball A is released from rest; ball B is launched horizontally at 8 m/s. They do not collide. Use fixed horizontal x and upward y, neglect air resistance, and take uniform $g=10\ \mathrm{m/s^2}$. After release, $\vec a=-g\hat\jmath$.

1. Find each impact time.

2. Find the horizontal separation of their impact points.

3. Compare their impact speeds and explain why equal fall times do not imply equal speeds.

Question 2 - Solution

A horizontal launch and a drop land together

(a) Solve the common vertical motion.

$$0=20-\tfrac12(10)t^2\ \Rightarrow\ \boxed{t_A=t_B=2\,\mathrm s}$$ Both balls have $v_{0y}=0$ and vertical position $y=20\ \mathrm m-\tfrac12gt^2$. Setting y to zero gives $t=\sqrt{2(20)/10}=2$ s for each. Horizontal launch speed does not enter this equation.

(b) Evolve the different horizontal motions.

$$\Delta x=x_B-x_A=(8-0)(2)=16\,\mathrm m$$ A remains over its release point. B travels $x=(8\ \mathrm{m/s})(2\ \mathrm s)=16$ m. Thus their impact points are 16 m apart despite identical vertical histories.

(c) Combine the impact components.

$$v_A=|0-10(2)|=20\,\mathrm{m/s},\quad v_B=\sqrt{8^2+20^2}=4\sqrt{29}\,\mathrm{m/s}$$ Both arrive with $v_y=-20$ m/s. A has speed 20 m/s; B has speed $\sqrt{8^2+20^2}=4\sqrt{29}\approx21.541$ m/s. Horizontal motion adds to B’s total speed but does not delay its vertical descent in this ideal model.