The problem involves solving for the time $t$ it takes for a projectile to fall to Earth using the height formula:

$h = -\frac{1}{2} g t^2 + v_0 t + h_0$

where:

• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity)
• $v_0 = 60 \, \text{m/s}$ (initial vertical velocity)
• $h_0 = 3 \, \text{m}$ (initial height).

When the projectile hits the ground, $h = 0$. The equation becomes:

$0 = -\frac{1}{2}(9.8)t^2 + 60t + 3$

Let’s solve for $t$. First, simplify the equation:

$-4.9t^2 + 60t + 3 = 0$

This is a quadratic equation in the form $at^2 + bt + c = 0$, where:

• $a = -4.9$
• $b = 60$
• $c = 3$.

The quadratic formula is:

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

Step-by-step solution:

1. Compute the discriminant $\Delta$: $\Delta = b^2 - 4ac = 60^2 - 4(-4.9)(3)$ $\Delta = 3600 + 58.8 = 3658.8$

2. Solve for $t$ using the quadratic formula: $t = \frac{-60 \pm \sqrt{3658.8}}{2(-4.9)}$ Compute $\sqrt{3658.8}$: $\sqrt{3658.8} \approx 60.49$

   Substitute back: $t = \frac{-60 \pm 60.49}{-9.8}$

3. Solve for the two roots:

   • $t_1 = \frac{-60 + 60.49}{-9.8} = \frac{-0.49}{-9.8} \approx 0.05 \, \text{s}$
   • $t_2 = \frac{-60 - 60.49}{-9.8} = \frac{-120.49}{-9.8} \approx 12.29 \, \text{s}$

4. Since $t > 0$, the time it takes for the projectile to fall to Earth is approximately:

$t \approx 12.29 \, \text{seconds}$

Would you like me to provide further explanation or verify any specific part?

Related Questions:

1. How would the solution change if $v_0 = 0$?
2. How do you derive the formula for $h$ using kinematic equations?
3. What happens to the projectile’s motion when air resistance is included?
4. Can you graph the height as a function of time?
5. What does the discriminant $\Delta$ tell us about the motion?

Tip:

When solving quadratic equations, always check the discriminant first to determine the nature of the roots!