Projectile Motion Analyzed — Question 10

Question 10

A three-dimensional launch remains in a vertical plane

Use z upward and horizontal x,y. A projectile leaves the origin with velocity $(6,8,15)$ m/s and acceleration $(0,0,-10)$ m/s$^2$. Neglect air resistance and consider flight until return to z=0.

1. Find flight time, landing coordinates and horizontal range.

2. Find the equation of the vertical plane containing the trajectory and the maximum height.

3. Does the projectile pass through $(12,16,10)$ m? Test all coordinates at one time.

Question 10 - Solution

A three-dimensional launch remains in a vertical plane

(a) Use vertical motion to set the time.

$$z=15t-5t^2=0\ \Rightarrow\ \boxed{T=3\,\mathrm s},\quad\vec r_f=(18,24,0)\,\mathrm m,\quad R=30\,\mathrm m$$ $z=15t-5t^2$ gives positive return time 3 s. Landing coordinates are $(18,24,0)$ m. Horizontal range is $\sqrt{18^2+24^2}=30$ m, also the horizontal speed $\sqrt{6^2+8^2}=10$ m/s times 3 s.

(b) Eliminate time from the horizontal pair.

$$\frac yx=\frac86=\frac43,\quad z_{\max}=\frac{15^2}{2(10)}=11.25\,\mathrm m$$ $x=6t$ and $y=8t$ imply $y=4x/3$. This plane extends vertically and contains the complete trajectory. Maximum z is $15^2/(2\cdot10)=11.25$ m at 1.5 s. Motion in space need not fill three independent spatial directions.

(c) Use a common candidate time.

$$t_x=12/6=2\,\mathrm s=t_y=16/8,\quad z(2)=15(2)-5(2)^2=10\,\mathrm m$$ Both x=12 m and y=16 m require t=2 s. At that time z equals $30-20=10$ m, so the projectile does pass through the target. A point outside $y=4x/3$ would be unreachable regardless of its height.