Projectile Motion Analyzed — Question 6

Question 6

An obstacle must be checked at its horizontal coordinate

A projectile leaves the origin with velocity $(10,15)$ m/s. A thin wall stands at x=20 m with top at y=8 m. It otherwise flies over level ground. 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 height and velocity when it reaches the wall, and the vertical clearance.

2. Find the greatest wall height it can just reach at that same horizontal position.

3. At what other horizontal position is the projectile at the same height as at x=20 m? Distinguish the two velocity directions.

Question 6 - Solution

An obstacle must be checked at its horizontal coordinate

(a) Use horizontal motion to determine the event time.

$$t_w=20/10=2\,\mathrm s,\quad y_w=15(2)-5(2)^2=10\,\mathrm m,\quad\boxed{\text{clearance}=10-8=2\,\mathrm m}$$ The wall is reached at $t=20/10=2$ s. Height is $15(2)-5(2)^2=10$ m and velocity is $(10,-5)$ m/s. The projectile is descending and clears the 8 m top by 2 m vertically.

(b) Hold the wall position fixed.

$$H_{{\mathrm{wall}},\max}=y(x=20\,\mathrm m)=10\,\mathrm m,\quad\vec v_w=(10,-5)\,\mathrm{m/s}$$ At x=20 m the only flight time is 2 s, so the limiting top height is 10 m. A top exactly there is grazed by the point-particle model; a higher wall blocks the trajectory. The maximum height elsewhere on the path is not the clearance at this wall.

(c) Solve the same-height equation.

$$10=15t-5t^2\ \Rightarrow\ t=1,2\,\mathrm s:\quad x_{\mathrm{other}}=10\,\mathrm m,\quad\vec v_{\mathrm{other}}=(10,5)\,\mathrm{m/s}$$ $10=15t-5t^2$ has roots 1 and 2 s. The other location is x=10 m, where velocity is $(10,5)$ m/s and motion is rising. Equal heights give opposite vertical velocities here, but different horizontal locations.