Testing a Launcher Under a Low Ceiling
A student team at a university open day is testing a small spring-loaded launcher in a covered hall. The launcher is fixed at ground level and fires a metal ball at 15 m/s at 55° above the horizontal. A solid ceiling runs from 6.0 m to 18.0 m horizontally from the launch point, at a height of 10.0 m; the ball cannot pass through it. At the far end of the hall, 18.0 m from the launcher, a target bullseye is mounted with its centre 5.0 m above the floor. The team wants to know whether the ball will clear the ceiling, whether it will hit the bullseye, and what minimum launch speed would be needed to hit the bullseye if the angle could be chosen freely. They assume air resistance is negligible and use g = 9.8 m/s². A teacher warns that increasing the speed while keeping 55° may solve one problem but create another, so the team must check the ceiling again for any proposed change.
1. Show whether the ball hits the ceiling before x = 18 m.
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Use the trajectory equation [y=x\tan\theta-\frac{gx^2}{2v^2\cos^2\theta}.] With (v=15), (\theta=55^\circ), (\tan55^\circ=1.428), and (\cos^2 55^\circ=0.3290), [y=1.428x-\frac{9.8}{2(225)(0.3290)}x^2 =1.428x-0.0662x^2.] At the start of the ceiling, (x=6.0): [y=1.428(6)-0.0662(36)=8.57-2.38=6.19\text{ m}.] This is below 10 m. The maximum height is [y_{\max}=\frac{(v\sin\theta)^2}{2g} =\frac{(15\sin55^\circ)^2}{19.6} =\frac{(12.29)^2}{19.6}\approx 7.70\text{ m}.] Since the highest point is below 10 m, the ball never reaches the ceiling. It clears the ceiling over the whole interval from 6 m to 18 m.
2. Determine whether the ball hits the bullseye at (18, 5). If not, calculate the minimum speed needed to hit that point, ignoring the ceiling.
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At (x=18), [y=1.428(18)-0.0662(18^2)=25.71-21.45=4.26\text{ m}.] The bullseye centre is at 5.0 m, so the ball passes about 0.74 m below it and misses.
The minimum speed needed to reach a point ((x,y)) from the origin is [v_{\min}^2=g\left(y+\sqrt{x^2+y^2}\right).] Here (x=18), (y=5), so [\sqrt{18^2+5^2}=\sqrt{349}=18.68,] [v_{\min}^2=9.8(5+18.68)=9.8(23.68)=232.1,] [v_{\min}=\sqrt{232.1}\approx 15.23\text{ m/s}.] The current speed is 15 m/s, so it is below the minimum by about 0.23 m/s. At least 15.23 m/s is required, whatever the angle.
3. The team raises the speed to 16.0 m/s but keeps 55°. Find where the ball crosses x = 18 and then determine the two possible launch angles at 16.0 m/s that would hit the target. Explain which angle is actually usable given the ceiling.
Read the worked answer
With (v=16), (\theta=55^\circ): [y=1.428x-\frac{9.8}{2(256)(0.3290)}x^2 =1.428x-0.0582x^2.] At (x=18): [y=25.71-0.0582(324)=25.71-18.85=6.86\text{ m}.] So the ball now passes about 1.86 m above the bullseye.
To hit ((18,5)) at 16 m/s, let (T=\tan\theta). The trajectory equation gives [5=18T-\frac{9.8(18^2)}{2(16^2)}(1+T^2) =18T-6.202(1+T^2).] Rearranging: [6.202T^2-18T+11.202=0.] The discriminant is [18^2-4(6.202)(11.202)=46.2,] so [T=\frac{18\pm\sqrt{46.2}}{12.404} =1.999\text{ or }0.903.] Thus (\theta\approx 63.4^\circ) or (42.1^\circ).
Check the ceiling. At (63.4^\circ), the vertical velocity is about (16\sin63.4^\circ=14.3) m/s, giving a maximum height of about (10.4) m. The ball reaches 10 m at roughly (x=8.3) m, which is under the ceiling, so it would hit the ceiling. At (42.1^\circ), the maximum height is only about (5.9) m, so it clears the ceiling. Therefore only the (42.1^\circ) launch is usable.