Question

which of the following would allow the cannonball to travel a greater horizontal distance, or range? increasing the launch speed all of these increasing the launch angle increasing the air resistance

Explanation:

Step1: Analyze the effect of launch speed

The range formula for projectile motion (assuming no air resistance initially) is \(R=\frac{v_0^{2}\sin2\theta}{g}\), where \(v_0\) is the launch speed. If we increase \(v_0\) (while keeping \(\theta\) constant), since \(R\propto v_0^{2}\), the range \(R\) will increase.

Step2: Analyze the effect of launch angle

The function \(y = \sin2\theta\) has a maximum value of 1 when \(2\theta = 90^{\circ}\) or \(\theta=45^{\circ}\). As we increase \(\theta\) from \(0^{\circ}\) to \(45^{\circ}\), \(\sin2\theta\) increases (so \(R\) increases). When \(\theta\) is in the range \(45^{\circ}<\theta < 90^{\circ}\), \(\sin2\theta\) decreases (because \(2\theta\) is in the range \(90^{\circ}<2\theta < 180^{\circ}\)). But if we assume we are increasing \(\theta\) within the range where \(\sin2\theta\) is still increasing (or considering appropriate angle - speed combinations), it can contribute to a larger range in some cases (in combination with speed, etc.). Also, if we consider the fact that for a given non - optimal speed, adjusting the angle can optimize the range.

Step3: Analyze the effect of air resistance

Air resistance \(F_d=-\frac{1}{2} ho v^{2}AC_d\hat{v}\) (where \( ho\) is the air density, \(v\) is the speed, \(A\) is the cross - sectional area, \(C_d\) is the drag coefficient) acts opposite to the direction of motion. It reduces both the horizontal and vertical components of velocity. Mathematically, the range formula in the presence of air resistance \(R=\frac{v_0^{2}\sin2\theta}{g + \frac{ ho v_0^{2}C_dA}{2m}}\) (a more complex form compared to the no - air - resistance case). Increasing air resistance (increasing \( ho\), \(C_d\), \(A\)) will decrease the range.

Since increasing launch speed and (in an appropriate range) increasing launch angle can increase the range (and increasing air resistance decreases the range), the option “ALL of these” is incorrect. But if we assume that the question is considering the positive - impact factors (increasing speed and in - range angle increase range, and maybe a mis - understanding in the problem setup), but actually, increasing air resistance is wrong. However, if we consider the standard projectile motion (ignoring air resistance for the positive - impact factors part), the range formula \(R=\frac{v_0^{2}\sin2\theta}{g}\). Increasing \(v_0\) (speed) makes \(R\) larger. And when \(\theta\) is such that \(\sin2\theta\) is maximized (or in a range where for a given non - optimal \(\theta\), increasing it towards the optimal \(45^{\circ}\) can help). But if we consider the options as: increasing launch speed (correct for range increase), increasing launch angle (if in the correct sub - range, correct), increasing air resistance (wrong). But if the problem is in a context where air resistance is ignored (a common physics problem simplification for basic range discussion), and we just consider the variables \(v_0\) and \(\theta\) (where increasing \(v_0\) always helps and increasing \(\theta\) in the right interval helps), and the option “ALL of these” is a trick. But actually, the correct factors for increasing range (ignoring air resistance) are increasing launch speed and adjusting (not just “increasing” in a non - bounded sense) the launch angle. But if we assume the problem is in a basic physics (projectile motion, no air resistance) and the options:
- Increasing launch speed: correct (\(R\propto v_0^{2}\))
- Increasing launch angle (assuming we increase it towards \(45^{\circ}\)): correct (since \(R = \frac{v_0^{2}\sin2\theta}{g}\) and \(\sin2\theta\) increases as \(\theta\) goes from \(0^{\circ}\) to \(45^{\circ}\))
- Increasing air resistance: wrong. But if the problem has an error and we consider only the first two (speed and in - range angle increase), but among the given options, if we assume that the question is in a very basic sense (maybe a misprint where air resistance is a distractor and the intended answer is based on speed and angle), but actually, the formula \(R=\frac{v_0^{2}\sin2\theta}{g}\) (no air resistance).

Answer:

Increasing the launch speed.