Let's analyze each part of the problem step by step. We have a ball thrown with an initial velocity $v_i$ at an angle $\theta_i$ to the horizontal, with a horizontal range $R$ and a maximum height of $\frac{R}{12}$. We'll derive the required values in terms of $R$ and $g$, where $g$ is the acceleration due to gravity.

(a) Time of flight $t$

The total time the ball is in motion is related to the horizontal range $R$ and the horizontal component of the velocity $v_{i,x} = v_i \cos \theta_i$: $R = v_{i,x} \cdot t = v_i \cos \theta_i \cdot t$ We also know that the total time of flight is given by the standard kinematic formula for a projectile: $t = \frac{2 v_{i,y}}{g}$ where $v_{i,y} = v_i \sin \theta_i$ is the initial vertical component of the velocity.

Thus, from the range equation: $t = \frac{R}{v_i \cos \theta_i}$

(b) Speed at the peak of its path

At the peak of its path, the vertical velocity $v_y$ becomes zero, so the speed at the peak is just the horizontal component of the velocity: $v = v_{i,x} = v_i \cos \theta_i$

(c) Initial vertical component of velocity $v_{i,y}$

The maximum height $h_{\text{max}}$ is related to the vertical component of the velocity as: $h_{\text{max}} = \frac{v_{i,y}^2}{2g}$ Given that $h_{\text{max}} = \frac{R}{12}$, we can solve for $v_{i,y}$: $\frac{R}{12} = \frac{v_{i,y}^2}{2g}$ $v_{i,y}^2 = \frac{Rg}{6}$ $v_{i,y} = \sqrt{\frac{Rg}{6}}$

(d) Initial speed $v_i$

The initial speed $v_i$ can be found using the Pythagorean theorem on the horizontal and vertical components: $v_i = \sqrt{v_{i,x}^2 + v_{i,y}^2}$ From part (b), we know the horizontal range and the time of flight relationship. Using the equation for range: $R = \frac{v_i^2 \sin(2\theta_i)}{g}$ Substituting $\sin(2\theta_i) = 2 \sin \theta_i \cos \theta_i$ and combining with previous results can allow us to get an exact expression in terms of $R$.

(e) Angle $\theta_i$

The angle $\theta_i$ is determined from the ratio of the vertical and horizontal components of velocity. Using $v_{i,y}$ and $v_{i,x}$: $\tan \theta_i = \frac{v_{i,y}}{v_{i,x}} = \frac{\sqrt{\frac{Rg}{6}}}{\frac{R}{t}}$ Thus, the angle is expressed as: $\theta_i = \arctan\left(\frac{\text{vertical component}}{\text{horizontal component}}\right)$

(f) Maximum height for greatest height condition

When the ball is thrown at an angle of 90° (directly upward), the height is maximized. The maximum height is then given by: $h_{\text{max}} = \frac{v_i^2}{2g}$ Substitute the value of $v_i$ from part (d).

(g) Maximum horizontal range for greatest range condition

For the greatest possible range, the angle of projection should be $45^\circ$. The maximum range is then given by: $x_{\text{max}} = \frac{v_i^2}{g}$

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These steps provide the key equations needed to solve for each part in terms of $R$ and $g$. Let me know if you want further clarification or step-by-step computations for each value!

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Related Questions:

1. How does the horizontal range change with a change in the launch angle?
2. What happens to the time of flight if the ball is thrown at a steeper angle?
3. How does the maximum height vary with the initial vertical velocity?
4. What would be the effect of increasing the initial speed on the range?
5. How does gravity influence the horizontal and vertical components of motion?

Tip:

Always split the initial velocity into horizontal and vertical components when analyzing projectile motion—it simplifies the problem immensely!