Topic: Kinematics
Concepts: Limiting cases, Projectile motion

Solution:

• A) This statement is true. Recall the range equation,
  
  $R=\frac{v^2\sin(2\theta)}{g}$
  
  which is maximized by $\theta=\pi/4$ so $\sin(2\theta)=1$,
  
  $R_{\text{max}}=\frac{v^2}{g}$
  
  If Tom can throw with maximum speed $v_{\text{max}}$, then we need
  
  $l \leq R_{\text{max}}=\frac{v^2_{\text{max}}}{g}$
  $v_{\text{max}} \geq \sqrt{gl}$
  
  for the ball to reach Wes.
• B) This statement is true. When $v_{\text{max}}=\sqrt{gl}$, there is only one trajectory which reaches Wes so $t_{\text{min}}=t_{\text{max}}=t_{\pi/4}$ where $t_{\pi/4}$ is the time taken by the trajectory with $\theta=\pi/4$. As $v_{\text{max}} \rightarrow \infty$, the window of times expands since at high speeds we can launch almost horizontally to reduce $t$ and almost vertically to increase $t$.

• C) This statement is true as explained in answer choice B.
• D) This statement is true. At small $l$, we have a range of times the ball could take to reach Wes depending on launch angle. As $l$ increases, this window closes since at the maximum $l=v^2_{\text{max}}/g$ there is only one trajectory which reaches Wes so $t_{\text{min}}=t_{\text{max}}=t_{\pi/4}$.

• E) This statement is false as explained in answer choice D.

Thus, the answer is E.