Projectile Motion Analyzed — Question 2

Question 2

A height measurement can correspond to two times

A projectile starts at the origin with velocity $(12,20)$ m/s. A camera records it at height 15 m. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. Find all possible recording times and horizontal positions.

2. Find its velocity and speed at each event.

3. If the camera records downward motion, select the event. What happens if the requested height is 25 m instead?

Question 2 - Solution

A height measurement can correspond to two times

(a) Retain both roots of the height equation.

$$15=20t-5t^2\ \Rightarrow\ (t-1)(t-3)=0,\quad\boxed{(t,x)=(1,12)\text{ or }(3,36)}\quad\text{(SI)}$$ $15=20t-5t^2$ gives $t^2-4t+3=0$, hence 1 s and 3 s. The horizontal positions are 12 and 36 m. A height by itself does not identify a unique point on the flight.

(b) Use vertical velocity to identify the branch.

$$\vec v(1)=(12,10),\quad\vec v(3)=(12,-10)\quad\mathrm{m/s},\quad v=\sqrt{12^2+10^2}=2\sqrt{61}\,\mathrm{m/s}$$ The velocities are $(12,10)$ and $(12,-10)$ m/s, each with speed $\sqrt{244}=2\sqrt{61}\approx15.620$ m/s. The first event is rising; the second is falling.

(c) Apply the additional observation and feasibility bound.

$$v_y<0\ \Rightarrow\ t=3\,\mathrm s,\quad H=\frac{20^2}{2(10)}=20\,\mathrm m<25\,\mathrm m$$ Downward motion selects 3 s and x=36 m. Maximum height is $v_{0y}^2/(2g)=20$ m, so 25 m is unreachable. Equivalently the height quadratic would have no real roots. Real roots must also lie within the physical airborne interval.