To determine how long after the pebble falls it will hit the ground, we need to solve for the time [tex]\( t \)[/tex] when the height [tex]\( h(t) \)[/tex] is 0. The height is given by the quadratic equation:
[tex]\[ h(t) = -4.9 t^2 + h_0 \][/tex]
From the problem, we are given that the quadratic equation that models the situation is:
[tex]\[ h(t) = -4.9 t^2 + 60 \][/tex]
where [tex]\( h_0 = 60 \)[/tex] meters is the initial height from which the pebble falls.
To find the time [tex]\( t \)[/tex] when the pebble hits the ground, set [tex]\( h(t) = 0 \)[/tex] and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -4.9 t^2 + 60 \][/tex]
Rearrange the equation to isolate [tex]\( t \)[/tex]:
[tex]\[ -4.9 t^2 + 60 = 0 \][/tex]
Add [tex]\( 4.9 t^2 \)[/tex] to both sides:
[tex]\[ 4.9 t^2 = 60 \][/tex]
Divide both sides by 4.9 to solve for [tex]\( t^2 \)[/tex]:
[tex]\[ t^2 = \frac{60}{4.9} \][/tex]
Calculate the right-hand side:
[tex]\[ t^2 \approx 12.2449 \][/tex]
Take the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ t \approx \sqrt{12.2449} \][/tex]
This gives us two solutions, [tex]\( t \approx 3.5 \)[/tex] and [tex]\( t \approx -3.5 \)[/tex]. Since time [tex]\( t \)[/tex] cannot be negative in this context, we discard the negative solution.
Therefore, the time [tex]\( t \)[/tex] to the nearest tenth of a second when the pebble hits the ground is:
[tex]\[ t \approx 3.5 \][/tex]
So, the pebble will hit the ground approximately 3.5 seconds after it falls.
