Let's break down the problem: - **Given:** - The stone is thrown upward from a bridge 20 m above water. - It hits the water 8 seconds later. - We are to find the speed with which it was thrown upward (\(v_i\)). - Assume \(g = 9.8 \, m/s^2\). Let's use the equation for vertical displacement: \[ y = v_i t - \frac{1}{2}gt^2 \] Here, \(y = -20\) m (since it ends up 20 m below the starting point), \(t = 8\) s, \(g = 9.8\) m/s². Plug in the values: \[ -20 = v_i (8) - \frac{1}{2}(9.8)(8^2) \] \[ -20 = 8v_i - 0.5 \times 9.8 \times 64 \] \[ -20 = 8v_i - 313.6 \] \[ 8v_i = 313.6 - 20 \] \[ 8v_i = 293.6 \] \[ v_i = \frac{293.6}{8} \] \[ v_i = 36.7 \text{ m/s} \] But this value is not among the options, so let's check the sign and calculations carefully. ### Rechecking: The equation should be: \[ y = v_i t + \frac{1}{2}(-g)t^2 \] \[ -20 = v_i \cdot 8 - 0.5 \cdot 9.8 \cdot (8)^2 \] \[ -20 = 8v_i - 313.6 \] \[ 8v_i = 293.6 \] \[ v_i = 36.7 \text{ m/s} \] Still, none of the answer choices match. Let’s check if the sign for the displacement should be positive (if we are taking downward as positive): \[ y = v_i t + \frac{1}{2}(-g)t^2 \] But since the stone ends up below the starting point, displacement is negative. All calculations are consistent, but the speed is not matching the given options. Let's check the **options** and see if there could be a negative sign. ### Let's try \(y = v_i t + \frac{1}{2} g t^2\) (if downward is positive): \[ 20 = v_i \cdot 8 + 0.5 \cdot 9.8 \cdot (8^2) \] \[ 20 = 8v_i + 313.6 \] \[ 8v_i = 20 - 313.6 = -293.6 \] \[ v_i = \frac{-293.6}{8} = -36.7 \text{ m/s} \] Still, not matching. ### Let's try another approach Let's check the possible error. The correct method is above, but the provided options are: - \(-19.57\) m/s - \(-23.52\) m/s - \(10.25\) m/s - \(15.42\) m/s Let's try the first option in the equation: \[ -20 = v_i \cdot 8 - 0.5 \cdot 9.8 \cdot 64 \] \[ -20 = 8v_i - 313.6 \] \[ 8v_i = 313.6 - 20 = 293.6 \] \[ v_i = 36.7 \text{ m/s} \] But the options are negative; perhaps the stone was thrown **downward**. If so: \[ y = v_i t + \frac{1}{2} g t^2 \] \[ -20 = v_i \cdot 8 + 0.5 \cdot 9.8 \cdot (8^2) \] \[ -20 = 8v_i + 313.6 \] \[ 8v_i = -20 - 313.6 = -333.6 \] \[ v_i = \frac{-333.6}{8} = -41.7 \text{ m/s} \] Still not matching. Let's try plugging in the options to see which fits: #### Try \(v_i = -19.57\): \[ -20 = (-19.57) \cdot 8 - 313.6 \] \[ -20 = -156.56 - 313.6 = -470.16 \] No. #### Try \(v_i = -23.52\): \[ -20 = (-23.52) \cdot 8 - 313.6 \] \[ -20 = -188.16 - 313.6 = -501.76 \] No. #### Try \(v_i = 10.25\): \[ -20 = 10.25 \cdot 8 - 313.6 \] \[ -20 = 82 - 313.6 = -231.6 \] No. #### Try \(v_i = 15.42\): \[ -20 = 15.42 \cdot 8 - 313.6 \] \[ -20 = 123.36 - 313.6 = -190.24 \] No. --- ### Final Review All calculations point to a higher positive value, which is not in the options. **However, based on the options, the closest solution using the correct equation would be:** \[ y = v_i t - \frac{1}{2}g t^2 \] \[ -20 = v_i (8) - 0.5 \cdot 9.8 \cdot 64 \] \[ -20 = 8v_i - 313.6 \] \[ 8v_i = 313.6 - 20 = 293.6 \] \[ v_i = 36.7 \text{ m/s} \] None of the choices match. There may be a typo in the question options, or the question expects the absolute value. So, **none of the given options is mathematically correct** based on the standard equations of motion. But if you are required to pick the closest option, **pick the highest positive value:** - **Option: \(v_i = 15.42\,\text{m/s}\)** But please note: **the actual correct answer is \(36.7\,\text{m/s}\)**. The options likely have an error.