The ball reached a maximum height of approximately 39 feet.
To find out how high the ball went, we'll first need to determine its initial velocity. We can use the following kinematic equation to do this:
y = y₀ + v₀t + (1/2)at²
where
- y is the final height of the ball
- y₀ is the initial height of the ball (3 feet in this case)
- v₀ is the initial velocity of the ball (what we want to find)
- t is the time the ball is in the air (3 seconds in this case)
- a is the acceleration due to gravity (approximately -32.2 ft/s², since it acts downward)
Since the ball is in the air for 3 seconds, it takes half that time (1.5 seconds) to reach its maximum height, and another 1.5 seconds to fall back to the initial height. When the ball reaches its maximum height, its final velocity (v) will be 0. We can use another kinematic equation to find the initial velocity (v₀):
v = v₀ + at
where
- v is the final velocity (0 ft/s at the maximum height)
- a is the acceleration due to gravity (-32.2 ft/s²)
- t is the time it takes to reach the maximum height (1.5 seconds)
Plugging in the values, we get:
0 = v₀ - (32.2 ft/s²)(1.5 s)
Solving for v₀, we find:
v₀ = (32.2 ft/s²)(1.5 s) = 48.3 ft/s
Now that we have the initial velocity, we can use the first kinematic equation to find the maximum height (y):
y = y₀ + v₀t + (1/2)at²
Plugging in the values, we get:
y = 3 ft + (48.3 ft/s)(1.5 s) - (1/2)(32.2 ft/s²)(1.5 s)²
y ≈ 3 ft + 72.45 ft - 36.225 ft
y ≈ 39.225 ft