(2) A youl it thiown. Up luard With aninitia Neiocity Ct 25 mathrm(~m) / mathrm(s) How high does it go before it Start5 to four back hard.

## Step 1: Identify the given values and the formula<br />### We are given the initial velocity \( v = 25 \, \text{m/s} \) and need to find the maximum height. The relevant equation is \( v_f^2 = v_i^2 + 2ad \), where \( v_f = 0 \) at the maximum height.<br />## Step 2: Substitute the known values into the equation<br />### Substitute \( v_f = 0 \), \( v_i = 25 \, \text{m/s} \), and \( a = -9.8 \, \text{m/s}^2 \) into the equation \( 0 = (25 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)d \).<br />## Step 3: Solve for \( d \)<br />### Rearrange the equation to solve for \( d \):<br />\[ 0 = 625 \, \text{m}^2/\text{s}^2 - 19.6 \, \text{m/s}^2 \cdot d \]<br />\[ 19.6 \, \text{m/s}^2 \cdot d = 625 \, \text{m}^2/\text{s}^2 \]<br />\[ d = \frac{625 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} \]<br />\[ d \approx 31.8877551 \, \text{m} \]