Question

which of the following would allow the cannonball to travel to a greater maximum height? increasing the launch velocity all of these increasing the initial launch height increasing the launch angle

Explanation:

Step1: Analyze the effect of launch velocity

The maximum height formula for projectile motion is \(h = \frac{(v_0\sin\theta)^2}{2g}+h_0\) (where \(v_0\) is launch velocity, \(\theta\) is launch angle, \(g\) is acceleration due to gravity, \(h_0\) is initial height). If we increase \(v_0\), since \(h\) is proportional to \(v_0^{2}\) (when \(\theta\) and \(h_0\) are constant), the maximum height increases.

Step2: Analyze the effect of initial launch height

From the formula \(h = \frac{(v_0\sin\theta)^2}{2g}+h_0\), if we increase \(h_0\) (while \(v_0\) and \(\theta\) are constant), the value of \(h\) (maximum height) will directly increase.

Step3: Analyze the effect of launch angle

The vertical - component of the initial velocity is \(v_{0y}=v_0\sin\theta\). As \(\theta\) increases from \(0^{\circ}\) to \(90^{\circ}\), \(\sin\theta\) increases. When \(v_0\) and \(h_0\) are constant, since \(h=\frac{(v_0\sin\theta)^2}{2g}+h_0\), \(h\) increases as \(\sin\theta\) increases (because \(h\) is a function of \(\sin^{2}\theta\)).

Answer:

B. All of these