Question 8 of 14 , Step 1 of 1 $6 / 14$ Correct An arrow is shot vertically upward from a platform $50 \mathrm{ft}$ high at a rate of $174 \mathrm{ft} / \mathrm{sec}$. When will the arrow hit the ground? Use the formula: $h=-16 t^{2}+v_{0} t+h_{0}$. (Round your answer to the nearest tenth.) Answer How to enter your answer (opens in new window)

Step 1 ：The problem is asking for the time when the arrow hits the ground. This means we need to find the time 't' when the height 'h' is zero. We can use the given formula and set h to 0, then solve for t. The initial velocity \(v_0\) is given as 174 ft/sec and the initial height \(h_0\) is given as 50 ft. We can substitute these values into the formula and solve for t.

Step 2 ：Substitute \(v_0 = 174\) ft/sec and \(h_0 = 50\) ft into the formula \(h = -16t^2 + v_0t + h_0\), we get \(0 = -16t^2 + 174t + 50\).

Step 3 ：Solving the quadratic equation, we get two solutions for \(t\): \(t = \frac{87}{16} - \frac{\sqrt{8369}}{16}\) and \(t = \frac{87}{16} + \frac{\sqrt{8369}}{16}\).

Step 4 ：We are interested in the positive solution as time cannot be negative. We also need to round the answer to the nearest tenth as per the question.

Step 5 ：So, the solution for \(t\) is \(t = \frac{87}{16} + \frac{\sqrt{8369}}{16}\).

Step 6 ：Final Answer: The arrow will hit the ground at approximately \(\boxed{10.9}\) seconds.