Projectile Motion Analyzed — Question 9

Question 9

A moving target requires a common place and time

A cart moves on y=0 with $x_C(t)=10\ \mathrm m+(5\ \mathrm{m/s})t$. A projectile launches from $(0,0)$ with $v_{0y}=15$ m/s. Its horizontal launch component may be chosen. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. For $v_{0x}=10$ m/s, determine whether the projectile hits the cart at its ground return.

2. At what earlier time do their x coordinates agree, and why is that not a collision?

3. Choose $v_{0x}$ so that a collision occurs when the projectile returns to ground.

Question 9 - Solution

A moving target requires a common place and time

(a) Find the vertical event before comparing horizontal positions.

$$T=2v_{0y}/g=3\,\mathrm s,\quad\boxed{x_P(T)-x_C(T)=30-(10+15)=5\,\mathrm m}$$ The positive return time is $T=2(15)/10=3$ s. At that instant projectile x is 30 m and cart x is 25 m. It misses the cart by 5 m horizontally.

(b) Check the second coordinate at the proposed encounter.

$$10t=10+5t\ \Rightarrow\ t=2\,\mathrm s,\quad y_P-y_C=15(2)-5(2)^2=10\,\mathrm m$$ $10t=10+5t$ gives $t=2$ s. The projectile is then at height $15(2)-5(2)^2=10$ m, whereas the cart remains at y=0. Equal x is only alignment in projection.

(c) Satisfy both event conditions together.

$$v_{0x}T=10+5T\ \Rightarrow\ v_{0x}=\frac{25}{3}\,\mathrm{m/s},\quad\vec r_{\mathrm{hit}}=(25,0)\,\mathrm m$$ The vertical return time remains 3 s because $v_{0y}$ is fixed. The cart is at 25 m then, so choose $v_{0x}=25/3\approx8.333$ m/s. Both objects meet at $(25,0)$ m at 3 s. The adjustment changes the launch speed and angle, not the vertical flight clock.