Answer: To solve this problem, we can use the equations of motion for projectile motion. Let's calculate the time of flight, horizontal distance, and maximum height of the ball.
Time of Flight:
The time of flight can be determined using the vertical motion equation:
- h = v₀y * t - (1/2) * g * t²
where:
h = initial height = 3 ft
v₀y = initial vertical velocity = v₀ * sin(θ)
v₀ = initial speed = 125 ft/sec
θ = launch angle = 40 degrees
g = acceleration due to gravity = 32.17 ft/sec² (approximate value)
We need to solve this equation for time (t). Rearranging the equation, we get:
- (1/2) * g * t² - v₀y * t + h = 0
Using the quadratic formula, t can be determined as:
- t = (-b ± √(b² - 4ac)) / (2a)
where:
- a = (1/2) * g
- b = -v₀y
- c = h
Plugging in the values, we have:
- a = (1/2) * 32.17 = 16.085
- b = -125 * sin(40) ≈ -80.459
- c = 3
Solving the quadratic equation for t, we get:
- t = (-(-80.459) ± √((-80.459)² - 4 * 16.085 * 3)) / (2 * 16.085)
- t ≈ 7.29 seconds
Therefore, the ball was in the air for approximately 7.29 seconds.
Horizontal Distance:
The horizontal distance traveled by the ball can be calculated using the horizontal motion equation:
where:
- d = horizontal distance
- v₀x = initial horizontal velocity = v₀ * cos(θ)
Plugging in the values, we have:
- v₀x = 125 * cos(40) ≈ 95.44 ft/sec
- t = 7.29 seconds
d = 95.44 * 7.29
d ≈ 694.91 feet
Therefore, the ball traveled approximately 694.91 feet horizontally.
Maximum Height:
The maximum height reached by the ball can be determined using the vertical motion equation:
- h = v₀y * t - (1/2) * g * t²
Using the previously calculated values:
- v₀y = 125 * sin(40) ≈ 80.21 ft/sec
- t = 7.29 seconds
Plugging in these values, we can calculate the maximum height:
h = 80.21 * 7.29 - (1/2) * 32.17 * (7.29)²
h ≈ 113.55 feet
Therefore, the ball reached a maximum height of approximately 113.55 feet.