To approach this problem, we will utilize the provided projectile motion model. The model used for height in projectile motion is given by:
[tex]\[ h(t) = -4.5 t^2 + v_0 t + h_0 \][/tex]
where:
- [tex]\( h(t) \)[/tex] is the height of the projectile at time [tex]\( t \)[/tex] seconds after its launch,
- [tex]\( v_0 \)[/tex] is the initial velocity in meters per second, set to [tex]\( 34.3 \)[/tex] m/s,
- [tex]\( h_0 \)[/tex] is the initial height, which is [tex]\( 11 \)[/tex] meters.
To determine how long it will take for the object to reach its maximum height, we will use the time at which the maximum height occurs. In a quadratic equation of the form [tex]\( at^2 + bt + c \)[/tex], the vertex, giving the maximum (or minimum) value, occurs at:
[tex]\[ t = \frac{-b}{2a} \][/tex]
Here, the equation [tex]\( h(t) = -4.5 t^2 + 34.3 t + 11 \)[/tex] has the coefficients:
- [tex]\( a = -4.5 \)[/tex]
- [tex]\( b = 34.3 \)[/tex]
We can calculate the time to reach maximum height using:
[tex]\[ t = \frac{-34.3}{2(-4.5)} \][/tex]
[tex]\[ t = \frac{-34.3}{-9} \][/tex]
[tex]\[ t \approx 3.8111 \text{ seconds} \][/tex]
Next, we need to find the maximum height [tex]\( h(t) \)[/tex] achieved at [tex]\( t = 3.8111 \)[/tex] seconds. Substituting [tex]\( t \)[/tex] into the height equation:
[tex]\[ h(3.8111) = -4.5 (3.8111)^2 + 34.3 (3.8111) + 11 \][/tex]
Calculating this, we get:
[tex]\[ h(3.8111) \approx -4.5 (14.52) + 130.79 + 11 \][/tex]
[tex]\[ h(3.8111) \approx -65.34 + 130.79 + 11 \][/tex]
[tex]\[ h(3.8111) \approx 76.36 \text{ meters} \][/tex]
Therefore, the time it will take for the object to reach its maximum height is approximately [tex]\( 3.8111 \)[/tex] seconds, and the maximum height achieved by the object is approximately [tex]\( 76.36 \)[/tex] meters.
