An object is launched vertically in the air at 49 meters per second from a 5-meter tall platform. Using the projectile motion model h(t)=-4.9t^2+v_(0)t+h_(0) where h(t) is the height of the projectile t sec after its departure, v_(0) is the initial velocity in meters per second, and h_(0) is the initial height in meters, determine how long it will take for the object to reach its maximum height What is the maximum height? The object will reach its maomum height in square seconds

Here's how to solve this problem:<br /><br />1. **Identify the knowns:**<br /> * Initial velocity ($v_0$) = 49 m/s<br /> * Initial height ($h_0$) = 5 m<br /> * Acceleration due to gravity (implied) = -9.8 m/s² (This is where the -4.9 in the formula comes from: -9.8/2 = -4.9)<br /><br />2. **Find the time to reach maximum height:**<br /> * The time to reach maximum height is when the vertical velocity is zero. The formula for vertical velocity is the derivative of the height function: $v(t) = -9.8t + v_0$.<br /> * Set $v(t) = 0$ and solve for $t$:<br /> 0 = -9.8t + 49<br /> 9.8t = 49<br /> t = 49 / 9.8<br /> t = 5 seconds<br /><br />3. **Find the maximum height:**<br /> * Substitute the time (t = 5 seconds) back into the original height equation:<br /> $h(5) = -4.9(5^2) + 49(5) + 5$<br /> $h(5) = -4.9(25) + 245 + 5$<br /> $h(5) = -122.5 + 245 + 5$<br /> $h(5) = 127.5$ meters<br /><br />**Answer:**<br /><br />The object will reach its maximum height in $\boxed{5}$ seconds. The maximum height is 127.5 meters.<br />