The problem involves finding the time when the object hits the ground, using the provided equation for its height:

$s = -4.9t^2 + 19.6t + 58.8$

The object hits the ground when $s = 0$. Substituting $s = 0$ into the equation:

$0 = -4.9t^2 + 19.6t + 58.8$

This is a quadratic equation of the form $ax^2 + bx + c = 0$. To solve for $t$, we use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = -4.9$, $b = 19.6$, and $c = 58.8$. Let's calculate the solution step by step:

1. Calculate the discriminant: $\Delta = b^2 - 4ac = (19.6)^2 - 4(-4.9)(58.8)$

2. Simplify further: $\Delta = 384.16 + 1150.08 = 1534.24$

3. Find the roots: $t = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-19.6 \pm \sqrt{1534.24}}{2(-4.9)}$

4. Simplify the square root: $\sqrt{1534.24} = 39.17$

5. Substitute into the formula: $t = \frac{-19.6 \pm 39.17}{-9.8}$

   Solve for each case:

   • For $t = \frac{-19.6 + 39.17}{-9.8}$: $t = \frac{19.57}{-9.8} \approx -2 \quad (\text{discard, as time cannot be negative})$

   • For $t = \frac{-19.6 - 39.17}{-9.8}$: $t = \frac{-58.77}{-9.8} \approx 6 \, \text{seconds}$

Thus, the object strikes the ground at 6 seconds.

Would you like a detailed breakdown of these steps or help with further related questions?

Related Questions:

1. How is the quadratic formula derived, and why does it work for solving such problems?
2. Can this problem be solved graphically? How would the curve look?
3. What would happen if the initial velocity was changed to 15 m/s?
4. How would air resistance affect the time of flight?
5. What is the object's velocity just before it hits the ground?

Tip:

Always verify your solutions by substituting the time values back into the original equation to ensure they satisfy the condition $s = 0$.