Hugo lies on top of a building throwing pennies straight down to the street below. The formula for the height H, that a penny falls is `H = Vt + 5t^2`, where V is the original velocity of the penny (how fast Hugo throws it when it leaves has hand) and t is equal to the time it takes to hit the ground. The building is 60 meters high, and Hugo throws the penny down at an initial speed of 20 meters per second. How long does it take for the penny to hit the ground?

Answer

To solve the problem, we need to find the time it takes for the penny to hit the ground using the given formula for height \( H \):

\[ H = Vt + 5t^2 \]

where:
- \( H \) is the height from which the penny is thrown (60 meters),
- \( V \) is the initial velocity (20 meters per second),
- \( t \) is the time in seconds.

### Step 1: Set up the equation

We know that the height \( H \) is 60 meters, and the initial velocity \( V \) is 20 meters per second. We can substitute these values into the equation:

\[ 60 = 20t + 5t^2 \]

### Step 2: Rearrange the equation

To solve for \( t \), we need to rearrange the equation so that all terms are on one side:

\[ 5t^2 + 20t - 60 = 0 \]

### Step 3: Simplify the equation

Next, we can simplify this equation by dividing all terms by 5:

\[ t^2 + 4t - 12 = 0 \]

### Step 4: Factor the quadratic equation

Now we need to factor the quadratic equation \( t^2 + 4t - 12 = 0 \). We are looking for two numbers that multiply to -12 and add to 4. These numbers are 6 and -2. Thus, we can write:

\[ (t + 6)(t - 2) = 0 \]

### Step 5: Solve for \( t \)

Setting each factor equal to zero gives us:

1. \( t + 6 = 0 \) → \( t = -6 \) (not valid since time cannot be negative)
2. \( t - 2 = 0 \) → \( t = 2 \)

### Conclusion

The time it takes for the penny to hit the ground is:

\[ \boxed{2 \text{ seconds}} \]