We need to analyze the given quadratic equation for the height \( s(t) = -4.9t^2 + 19.6t + 58.8 \).
a. Initial Height
The height above the ground when the object is launched corresponds to \( s(0) \): \[ s(0) = -4.9(0)^2 + 19.6(0) + 58.8 = 58.8 \text{ meters} \]
b. Time Until Object Hits the Ground
To find the time when the object hits the ground, we set \( s(t) = 0 \): \[ -4.9t^2 + 19.6t + 58.8 = 0 \]
We can solve this quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -4.9 \), \( b = 19.6 \), and \( c = 58.8 \).
Calculating the discriminant: \[ b^2 - 4ac = (19.6)^2 - 4(-4.9)(58.8) \] \[ = 384.16 + 1155.84 = 1540 \]
Now applying the quadratic formula: \[ t = \frac{-19.6 \pm \sqrt{1540}}{2(-4.9)} \] Calculating \( \sqrt{1540} \approx 39.24 \): \[ t = \frac{-19.6 \pm 39.24}{-9.8} \]
Now, calculating the two potential solutions: 1. \[ t_1 = \frac{-19.6 + 39.24}{-9.8} = \frac{19.64}{9.8} \approx 2.00 \text{ seconds} \] 2. \[ t_2 = \frac{-19.6 - 39.24}{-9.8} = \frac{-58.84}{-9.8} \approx 5.99 \text{ seconds} (not valid because it doesn't give a positive height) \]
Thus, the object hits the ground at approximately 2.00 seconds after launch.
c. Maximum Height and Time to Reach It
The maximum height of the object occurs at the vertex of the parabola described by the quadratic equation. The vertex \( t \) for a quadratic equation in the form \( at^2 + bt + c \) is given by: \[ t = -\frac{b}{2a} \] Here \( a = -4.9 \) and \( b = 19.6 \): \[ t = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2 \text{ seconds} \]
Now we will substitute \( t = 2 \) back into the height equation to find the maximum height: \[ s(2) = -4.9(2^2) + 19.6(2) + 58.8 \] \[ = -4.9(4) + 39.2 + 58.8 \] \[ = -19.6 + 39.2 + 58.8 = 78.4 \text{ meters} \]
Summary
a. The height above the ground when the object is launched is 58.8 meters.
b. The time before the object hits the ground after launch is approximately 2.00 seconds.
c. The maximum height reached is 78.4 meters, and it occurs at 2 seconds.