Catapults date from thousands of years ago, and were used historically to launch

Question

Catapults date from thousands of years ago, and were used historically to launch everything from stones to horses. During a battle in what is now Bavaria, inventive artillerymen from the united German clans launched giant spaetzle from their catapults toward a Roman fortification whose walls were 8.50 m high. The catapults launched spaetzle projectiles from a height of 4.40 m above the ground, and a distance of 38.9 m from the walls of the fortification at an angle of 60.0 degrees above the horizontal (see figure). The projectiles were to hit the top of the wall, splattering the Roman soldier atop the wall with pulverized pasta. (For the following questions, ignore any effects due to air resistance.)

a)What launch speed was necessary? ans in m/s

b)How long were the spaetzle in the air? ans in m/s

c)At what speed did the projectiles hit the wall? ans in m/s

Explanation / Answer

a)

consider the motion along the X-direction :

X = horizontal displacement = 38.9 m

Vo = speed of launch

Vox = initial velocity along X-axis = Vo Cos60

t = time

using the equation

X = Vox t

X = Vo Cos60 t

t = 77.8/Vo eq-1

Along the y-direction

Yo = initial position = 4.40 m

Y = final position = 8.50 m

Voy = initial velocity along Y-axis = Vo Sin60

a = acceleration = - 9.8

using the equation

Y = Yo + Voy t + (0.5) a t2

8.50 = 4.40 + (Vo Sin60) (77.8/Vo) + (0.5) (-9.8) (77.8/Vo)2

Vo = 21.65 m/s

b)

using eq-1

t = 77.8/Vo = 77.8/21.65 = 3.6 sec

c)

along the Y-direction

Vfy = Voy + at

Vfy = (21.65 Sin60) + (-9.8) (3.6)

Vfy = - 16.5 m/s

along X-direction :

Vfx = Vox = 21.65 Cos60 = 10.825 m/s

speed while hitting the wall is given as

Vf = sqrt(Vfx2 + Vfy2) = sqrt((10.825)2 + (- 16.5)2) = 19.7 m/s

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