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Ballista

A ballista (an ancient missile weapon) fires a heavy dart at an angle of 45 degrees to the horizontal with an initial velocity of $60 m s^{-1}$ .

What is the maximum horizontal distance from a $12 m$ -high castle wall that the ballista can be deployed so that the darts go over the wall? The height from which the darts are fired is level with the foot of the wall.

Assume air resistance is negligible and that the darts can be modelled as particles. Take $g = 9.8 m s^{-2}$ .

(1)

340 m

Incorrect

Using the trajectory equation $y = x \tan θ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ}$ and substituting in $θ = 45^{\circ}$ , $v_{0} = 60 m s^{-1}$ and $y = 12 m$ , you get $12 = x \tan 45^{\circ} - \frac{9.8 x^{2}}{7200 \cos^{2} 45^{\circ}}$ .

By cancelling and rearranging you can get your equation into the form of the following quadratic: $9.8 x^{2} - 3600 x + 43200 = 0$ which gives you $x = 350 m$ or $x = 12 m$ and as you’re looking for the maximum distance you select $x = 350 m$ .

(2)

350 m

Correct

Using the trajectory equation $y = x \tan θ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ}$ and substituting in $θ = 45^{\circ}$ , $v_{0} = 60 m s^{-1}$ and $y = 12 m$ , you get $12 = x \tan 45^{\circ} - \frac{9.8 x^{2}}{7200 \cos^{2} 45^{\circ}}$ .

By cancelling and rearranging you can get your equation into the form of the following quadratic: $9.8 x^{2} - 3600 x + 43200 = 0$ which gives you $x = 350 m$ or $x = 12 m$ and as you’re looking for the maximum distance you select $x = 350 m$ .

(3)

380 m

Incorrect

Using the trajectory equation $y = x \tan θ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ}$ and substituting in $θ = 45^{\circ}$ , $v_{0} = 60 m s^{-1}$ and $y = 12 m$ , you get $12 = x \tan 45^{\circ} - \frac{9.8 x^{2}}{7200 \cos^{2} 45^{\circ}}$ .

By cancelling and rearranging you can get your equation into the form of the following quadratic: $9.8 x^{2} - 3600 x + 43200 = 0$ which gives you $x = 350 m$ or $x = 12 m$ and as you’re looking for the maximum distance you select $x = 350 m$ .

(4)

390 m

Incorrect

Using the trajectory equation $y = x \tan θ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ}$ and substituting in $θ = 45^{\circ}$ , $v_{0} = 60 m s^{-1}$ and $y = 12 m$ , you get $12 = x \tan 45^{\circ} - \frac{9.8 x^{2}}{7200 \cos^{2} 45^{\circ}}$ .

By cancelling and rearranging you can get your equation into the form of the following quadratic: $9.8 x^{2} - 3600 x + 43200 = 0$ which gives you $x = 350 m$ or $x = 12 m$ and as you’re looking for the maximum distance you select $x = 350 m$ .

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