Answer :
To determine the maximum height reached by the ball, we start with the given initial conditions and the equation for height as a function of time:
[tex]\[ h(t) = -16t^2 + v_0 t + h_0 \][/tex]
where [tex]\( v_0 = 110 \)[/tex] feet per second (initial velocity) and [tex]\( h_0 = 2 \)[/tex] feet (initial height).
To find the maximum height, we need to determine the time when the ball reaches its maximum height. This occurs when the velocity [tex]\( v(t) \)[/tex] is zero. The velocity function can be obtained by differentiating the height function with respect to time:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 110t + 2] = -32t + 110 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex] to find the time [tex]\( t \)[/tex] at which the maximum height is reached:
[tex]\[ -32t + 110 = 0 \][/tex]
[tex]\[ 32t = 110 \][/tex]
[tex]\[ t = \frac{110}{32} \][/tex]
[tex]\[ t = 3.4375 \, \text{seconds} \][/tex]
Now, plug the time [tex]\( t = 3.4375 \)[/tex] back into the height equation to find the maximum height:
[tex]\[ h(3.4375) = -16(3.4375)^2 + 110(3.4375) + 2 \][/tex]
[tex]\[ h(3.4375) = -16(11.8164) + 377.125 + 2 \][/tex]
[tex]\[ h(3.4375) = -189.0625 + 377.125 + 2 \][/tex]
[tex]\[ h(3.4375) = 190.0625 \, \text{feet} \][/tex]
Rounded to the nearest hundredth, the maximum height the ball will attain is:
[tex]\[ 191.06 \, \text{feet} \][/tex]
Therefore, the maximum height the ball will attain is [tex]\( \boxed{191.06} \, \text{feet} \)[/tex].
[tex]\[ h(t) = -16t^2 + v_0 t + h_0 \][/tex]
where [tex]\( v_0 = 110 \)[/tex] feet per second (initial velocity) and [tex]\( h_0 = 2 \)[/tex] feet (initial height).
To find the maximum height, we need to determine the time when the ball reaches its maximum height. This occurs when the velocity [tex]\( v(t) \)[/tex] is zero. The velocity function can be obtained by differentiating the height function with respect to time:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 110t + 2] = -32t + 110 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex] to find the time [tex]\( t \)[/tex] at which the maximum height is reached:
[tex]\[ -32t + 110 = 0 \][/tex]
[tex]\[ 32t = 110 \][/tex]
[tex]\[ t = \frac{110}{32} \][/tex]
[tex]\[ t = 3.4375 \, \text{seconds} \][/tex]
Now, plug the time [tex]\( t = 3.4375 \)[/tex] back into the height equation to find the maximum height:
[tex]\[ h(3.4375) = -16(3.4375)^2 + 110(3.4375) + 2 \][/tex]
[tex]\[ h(3.4375) = -16(11.8164) + 377.125 + 2 \][/tex]
[tex]\[ h(3.4375) = -189.0625 + 377.125 + 2 \][/tex]
[tex]\[ h(3.4375) = 190.0625 \, \text{feet} \][/tex]
Rounded to the nearest hundredth, the maximum height the ball will attain is:
[tex]\[ 191.06 \, \text{feet} \][/tex]
Therefore, the maximum height the ball will attain is [tex]\( \boxed{191.06} \, \text{feet} \)[/tex].
