To solve the problem, let's break it down step-by-step using the given equation for the height of a ball in projectile motion:
[tex]\[ h(t) = a t^2 + v t + h_0 \][/tex]
Where:
- [tex]\( h(t) \)[/tex] is the height of the ball at time [tex]\( t \)[/tex],
- [tex]\( a \)[/tex] is the acceleration due to gravity, which is [tex]\(-16 \, \text{ft/s}^2 \)[/tex] (since gravity acts downward),
- [tex]\( v \)[/tex] is the initial upward velocity, which is [tex]\( 36 \, \text{ft/s} \)[/tex],
- [tex]\( h_0 \)[/tex] is the initial height of the ball, which is [tex]\( 4 \, \text{ft} \)[/tex],
- [tex]\( t \)[/tex] is the time after the ball is thrown, in seconds. In this case, [tex]\( t = 2 \)[/tex] seconds.
Let's plug in the values into the formula:
1. Initial height: [tex]\( h_0 = 4 \, \text{ft} \)[/tex]
2. Initial velocity: [tex]\( v = 36 \, \text{ft/s} \)[/tex]
3. Acceleration due to gravity: [tex]\( a = -16 \, \text{ft/s}^2 \)[/tex]
4. Time: [tex]\( t = 2 \)[/tex] seconds
Now substitute these values into the equation:
[tex]\[ h(2) = (-16) (2)^2 + (36) (2) + 4 \][/tex]
Calculate each term step by step:
1. [tex]\( (-16) (2)^2 = (-16) (4) = -64 \)[/tex]
2. [tex]\( (36) (2) = 72 \)[/tex]
Now, add these results along with the initial height:
[tex]\[ h(2) = -64 + 72 + 4 \][/tex]
Perform the addition:
[tex]\[ h(2) = 12 \][/tex]
Therefore, the height of the ball 2 seconds after it is thrown is:
[tex]\[ \boxed{12 \, \text{ft}} \][/tex]
