6 An object is thrown vertically upward with an initial velocity of 9.81 meters per second . What is the maximum height reached by the object? [Neglect friction.] (1) 1.00 m (3) 9.81 m (2) 4.91 m (4) 19.6 m

To solve this problem, we can use the **kinematic equation** for vertical motion under constant acceleration due to gravity:<br /><br />\[<br />v^2 = u^2 + 2as<br />\]<br /><br />Where:<br />- \( v \) is the final velocity (at the maximum height, \( v = 0 \)),<br />- \( u \) is the initial velocity (\( u = 9.81 \, \text{m/s} \)),<br />- \( a \) is the acceleration (\( a = -9.81 \, \text{m/s}^2 \), since gravity acts downward),<br />- \( s \) is the displacement or height.<br /><br />### Step 1: Substitute known values into the equation<br />At maximum height, \( v = 0 \). The equation becomes:<br /><br />\[<br />0 = (9.81)^2 + 2(-9.81)s<br />\]<br /><br />Simplify:<br /><br />\[<br />0 = 96.2361 - 19.62s<br />\]<br /><br />### Step 2: Solve for \( s \)<br />Rearrange the equation to isolate \( s \):<br /><br />\[<br />19.62s = 96.2361<br />\]<br /><br />\[<br />s = \frac{96.2361}{19.62}<br />\]<br /><br />\[<br />s = 4.91 \, \text{m}<br />\]<br /><br />### Final Answer:<br />The maximum height reached by the object is:<br /><br />\[<br />\boxed{4.91 \, \text{m}}<br />\]<br /><br />Correct choice: **(2) 4.91 m**