Projectile Motion — Question 9

Question 9

A falling target shares the same gravitational drop

A projectile is aimed directly from the origin toward a target initially at $(12,9)$ m. Its launch speed is 15 m/s. At release the target drops from rest. Both then accelerate downward at g and may meet before reaching the floor y=0. 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 the projectile’s initial velocity components.

2. Find the time and position of interception using the relative position.

3. Check that the interception occurs above the floor. State an assumption on which cancellation of gravity depends.

Question 9 - Solution

A falling target shares the same gravitational drop

(a) Aim along the initial line of sight.

$$\vec v_0=15\frac{(12,9)}{\sqrt{12^2+9^2}}=\boxed{(12,9)\,\mathrm{m/s}}$$ The target’s initial separation is $\sqrt{12^2+9^2}=15$ m. Multiplying its unit direction $(12/15,9/15)$ by 15 m/s gives $\vec v_0=(12,9)$ m/s.

(b) Subtract histories at the same time.

$$\vec r_{T/P}=(12,9)-(12,9)t=\vec0\ \Rightarrow\ t=1\,\mathrm s,\quad\vec r=(12,4)\,\mathrm m$$ Target position is $(12,9)-5t^2\hat\jmath$ m and projectile position is $(12t,9t)-5t^2\hat\jmath$ m for numerical seconds. Their difference is $(12,9)-(12,9)t$. It vanishes at 1 s. Both positions are then $(12,4)$ m.

(c) Check the physical flight interval.

$$y(1)=9-5=4\,\mathrm m>0,\quad\vec a_{T/P}=\vec a_T-\vec a_P=\vec0$$ The meeting height is 4 m, so it precedes floor contact. The common drop cancels only because both objects begin falling at the same time and have the same uniform gravitational acceleration with no additional differential effects such as air drag.