Question 4 5 pts Imani throws a ball at some angle $\theta$ above the horizontal with initial speed 14.8 $\mathrm{m} / \mathrm{s}$. What will be the ball's speed when it reaches a height $2.26 \mathrm{~m}$ above the point where it left her hand? Assume $0^{\circ}<\theta<90^{\circ}$, air resistance is negligible, and the acceleration due to gravity is $a_{y}=-g$. Cannot be determined from the information given $0 \mathrm{~m} / \mathrm{s}$ $14.8 \mathrm{~m} / \mathrm{s}$ $13.2 \mathrm{~m} / \mathrm{s}$

Step 1 ：Given that the initial speed of the ball, \(v_{initial}\), is 14.8 m/s, the height, \(h\), is 2.26 m, and the acceleration due to gravity, \(g\), is 9.8 m/s^2.

Step 2 ：Calculate the initial kinetic energy of the ball, \(KE_{initial}\), using the equation \(KE_{initial} = \frac{1}{2} m v_{initial}^2\). Assuming the mass of the ball, \(m\), to be 1 kg, we get \(KE_{initial} = \frac{1}{2} \times 1 \times (14.8)^2 = 109.52\) J.

Step 3 ：Calculate the potential energy of the ball at the given height, \(PE\), using the equation \(PE = m g h\). We get \(PE = 1 \times 9.8 \times 2.26 = 22.148\) J.

Step 4 ：Calculate the final kinetic energy of the ball, \(KE_{final}\), using the equation \(KE_{final} = KE_{initial} - PE\). We get \(KE_{final} = 109.52 - 22.148 = 87.372\) J.

Step 5 ：Calculate the speed of the ball at the given height, \(v_{final}\), from the final kinetic energy using the equation \(v_{final} = \sqrt{\frac{2 KE_{final}}{m}}\). We get \(v_{final} = \sqrt{\frac{2 \times 87.372}{1}} = 13.219\) m/s.

Step 6 ：\(\boxed{\text{The ball's speed when it reaches a height of 2.26 m above the point where it left her hand is approximately 13.2 m/s.}}\)