Projectile Motion Analyzed — Question 1

Question 1

An elevated launch breaks the equal-height flight-time shortcut

A projectile leaves $(0,20)$ m with velocity $(8,15)$ m/s and lands on y=0. 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 landing time, explaining the rejected root.

2. Find the horizontal range and impact velocity.

3. Compare ascent time with the time from apex to landing. Why are they unequal?

Question 1 - Solution

An elevated launch breaks the equal-height flight-time shortcut

(a) Set the final height, not the displacement, to zero.

$$0=20+15T-5T^2=-5(T-4)(T+1)\ \Rightarrow\ \boxed{T=4\,\mathrm s}\quad(T\ge0)$$ $0=20+15t-5t^2$ for t in seconds. Factoring gives $(t-4)(t+1)=0$, so the physical landing is at 4 s. The root $-1$ s is a backward extrapolation before launch.

(b) Evaluate both components at impact.

$$R=8(4)=32\,\mathrm m,\quad\vec v_f=(8,15-40)=(8,-25)\,\mathrm{m/s},\quad v_f=\sqrt{689}\,\mathrm{m/s}$$ Range is $8(4)=32$ m. The impact velocity is $(8,15-40)=(8,-25)$ m/s, speed $\sqrt{689}\approx26.249$ m/s. The horizontal component remains unchanged throughout flight.

(c) Use the apex as an intermediate event.

$$t_\uparrow=v_{0y}/g=1.5\,\mathrm s,\quad t_\downarrow=T-t_\uparrow=2.5\,\mathrm s$$ The apex occurs at 1.5 s, so ascent takes 1.5 s and descent to the floor takes 2.5 s. Landing is 20 m below release, so the descending part covers extra vertical distance. The equal-height formula $T=2v_{0y}/g=3$ s would stop at the release height, not the floor.