Projectile Motion Analyzed — Question 5

Question 5

Infer the launch from two timed observations

The launch occurs at time zero with x=0 but unknown initial y. A projectile is observed at $(10,14)$ m at 1 s and $(30,4)$ m at 3 s. Neglect air resistance and use uniform $g=10\ \mathrm{m/s^2}$ downward. Unless stated otherwise, x is horizontal and y upward.

1. Find the initial horizontal and vertical velocity components and initial height.

2. Find the apex time and maximum height above y=0.

3. Predict the position at 2 s and explain why linearly interpolating the observed heights is incorrect.

Question 5 - Solution

Infer the launch from two timed observations

(a) Solve the component equations simultaneously.

$$v_{0x}=\frac{30-10}{3-1}=10,\quad y_0+v_{0y}=19,\quad y_0+3v_{0y}=49\ \Rightarrow\ \boxed{v_{0y}=15,\ y_0=4}\quad\text{(SI)}$$ Horizontal data give $v_{0x}=10$ m/s. The vertical records give $y_0+v_{0y}(1\ \mathrm s)=19$ m and $y_0+v_{0y}(3\ \mathrm s)=49$ m. Subtracting yields $v_{0y}=15$ m/s, then $y_0=4$ m.

(b) Measure the rise from the actual release height.

$$t_a=15/10=1.5\,\mathrm s,\quad y_a=4+\frac{15^2}{20}=15.25\,\mathrm m$$ The apex occurs at $15/10=1.5$ s. Its height is $4+15^2/(2\cdot10)=15.25$ m. The rise is 11.25 m, different from its absolute height above the chosen zero.

(c) Use the curved time dependence.

$$\vec r(2)=(20,4+30-20)=(20,14)\,\mathrm m,\quad14-\frac{14+4}{2}=5\,\mathrm m$$ At 2 s, $x=20$ m and $y=4+30-20=14$ m. Linear interpolation between the two heights would give 9 m. The vertical position is quadratic in time, so the midpoint lies 5 m above the endpoint secant for this 2 s interval.