To find the maximum height and initial velocity of the cricket ball, we can use the equations of motion for vertical motion.
Let's assume the upward direction is positive.
The time taken for the ball to reach its highest point is half of the total time of flight, which is 6 seconds divided by 2, giving us 3 seconds.
At the highest point, the vertical velocity of the ball becomes zero since it momentarily comes to rest before reversing its direction.
Using the equation for vertical displacement (assuming no air resistance):
Δy = V₀t + (1/2)gt²
where: Δy = change in height (max height) V₀ = initial vertical velocity t = time taken to reach max height (3 seconds) g = acceleration due to gravity (approximately 9.8 m/s²)
At the maximum height, the displacement is zero because the ball comes to rest before falling back down. Thus, we can rewrite the equation as:
0 = V₀(3) + (1/2)(9.8)(3)²
Simplifying the equation:
0 = 3V₀ + 44.1
Rearranging the equation:
3V₀ = -44.1
V₀ = -44.1/3
V₀ ≈ -14.7 m/s (upward)
Therefore, the initial velocity of the cricket ball is approximately 14.7 m/s in the upward direction.
To find the maximum height, we can substitute the value of V₀ into the equation for displacement:
Δy = (-14.7)(3) + (1/2)(9.8)(3)²
Simplifying the equation:
Δy = -44.1 + 44.1
Δy = 0
Therefore, the maximum height reached by the cricket ball is 0 meters.