Question

Suppose you want to throw a dart across a room at a wall-mounted target and hit the target exactly at its center point. We call this center point of the target its “bull’s-eye”. The bull’s-eye is a horizontal distance d from the dart’s point of release and a height habove the dart’s point of release. To hit the bull’s-eye, you have a choice of the speed v at which you throw the dart and a choice of the angle θ above the horizontal direction at which you release the dart. In other words, θ is the angle between the dart’s initial velocity vector and the horizontal direction. For any angle at which you choose to throw the dart, it can be shown that there is at most one speed that will allow it to hit the bull’s-eye. If you throw the dart too softly, it will land under the bull’s-eye, and if you throw the dart too hard, it will land over the bull’s-eye.

1.1. Draw a diagram of the scenario described above. Label the distances d and h and the angle θ on the diagram so that it’s clear what they mean, and draw roughly what the trajectory of the dart would look like if it were to hit the bull’s-eye.

1.2. For any given angle θ there is at most one speed at which you can throw the dart to hit the bull’s-eye, but is there also always at least one such speed, or are there angles at which there is no speed that will allow the dart to hit the bull’s-eye? Circle any of the following mathematical relationships that must have been satisfied by d, h and θ if the dart were to hit the bull’s eye. If you believe no such restriction exists, circle NONE. Explain you reasoning in the box below the answer choices.

tanθ < h/d tanθ = h/d   tanθ > h/d NONE

1.3. Which of the following is the correct expression for the speed at which you have to shoot the ball to make the shot? Circle the correct expression, and in the box below, explain your reasoning and/or any calculations which led to your choice.

cos θ cos θ |d tan θ-h cos θ Intan θ_d cos θ cos θ | h tan θ-d cos θ cos θ h tan θ-d 11 tan θ--d ldtan θ + h cos θ cos θ | d tan θ + h cos θ cos θ d tan θ + h dtan θ + h d htand cos θ cos θ cos θ | h tan θ + d cos θ h tan θ + d tan θ + d 2

Answer 1

There doesn't have to be atleast one speed, this will be clear from the velocity expression in the next part. For angles like 90 degree or 0 it is easy to see, that given any speed will not make the dart reach its target. If the launch point is the origin, then the equation of trajectory is y = x tan theta (1 - x/R) where R is the horizontal range of the projectile. To hit the bull's eye the point (x, y) = (d, h) must satisfy this equation. Thus, h = d tan theta (1 - d/R) R = d/1 - h/d tan theta u^2 sin 2 theta/g = d^2 tan theta/d tan theta - h u = squareroot d^2 g/2 cos^2 theta (d tan theta - h) = d/cos theta squareroot g/2/d tan theta - h So that the condition for real speed is d tan theta - h > 0

There doesn't have to be atleast one speed, this will be clear from the velocity expression in the next part. For angles like 90 degree or 0 it is easy to see, that given any speed will not make the dart reach its target.

If the launch point is the origin, then the equation of trajectory is

$y=x\tan\theta\left ( 1-\frac{x}{R} \right )$

where R is the horizontal range of the projectile.

To hit the bull's eye the point $(x,y)=(d,h)$ must satisfy this equation. Thus,

$h=d\tan\theta\left ( 1-\frac{d}{R} \right )$

$R=\frac{d}{1-\frac{h}{d\tan\theta}}$

$\frac{u^2\sin2\theta}{g}=\frac{d^2\tan\theta}{d\tan\theta-h}$

$u=\sqrt{\frac{d^2g}{2\cos^2\theta(d\tan\theta-h)}}=\frac{d}{\cos\theta}\sqrt{\frac{g/2}{d\tan\theta-h}}$

So that the condition for real speed is

$d\tan\theta-h>0$