To solve the equation \( h = -16t^2 + 60t + 0 \), we can factor or use the quadratic formula to find the values of \( t \) when \( h = 0 \).
Thus, we need to solve: \[ -16t^2 + 60t = 0 \]
We can factor out \( t \): \[ t(-16t + 60) = 0 \]
Setting each factor to zero gives us:
- \( t = 0 \)
- \( -16t + 60 = 0 \)
For the second equation: \[ -16t + 60 = 0 \ 16t = 60 \ t = \frac{60}{16} \ t = \frac{15}{4} \ t = 3.75 \]
Thus, the solutions to the equation \( h = 0 \) are: \[ t = 0 \quad \text{and} \quad t = 3.75 \]
Since the initial question asked for the solution to the equation \( h = -16t^2 + 60t + 0 \) with rounding to the nearest hundredth, the roots are:
\[ t = 0.00 \quad \text{and} \quad t = 3.75 \]