Final answer:
The maximum height the ball can reach is 65.75 m. The ball will strike the ground approximately 3.696 seconds after being projected upward. The ball will be moving at approximately -0.247 m/s just before it hits the ground.
Step-by-step explanation:
To solve this problem, we can use the equations of motion for a projectile. Since the initial velocity is only in the vertical direction, we can separate the motion into vertical and horizontal components.
(a) Maximum height:
The maximum height can be found using the equation:
vf^2 - vi^2 = 2ad
Where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (36.2 m/s), a is the acceleration (-9.8 m/s^2), and d is the displacement in the vertical direction (the maximum height).
By substituting the known values into the equation, we can solve for d:
0 - (36.2)^2 = 2*(-9.8)*d
d = 65.75 m
Therefore, the maximum height the ball can reach is 65.75 m.
(b) Time to strike the ground:
The time it takes for the ball to strike the ground can be calculated using the equation:
t = (vf - vi) / a
Where vf is the final velocity (0 m/s), vi is the initial velocity (36.2 m/s), and a is the acceleration (-9.8 m/s^2).
By substituting the known values into the equation, we can solve for t:
t = (0 - 36.2) / (-9.8)
t ≈ 3.696 s
Therefore, the ball strikes the ground approximately 3.696 seconds after being projected upward.
(c) Velocity just before hitting the ground:
The velocity just before the ball hits the ground can be calculated using the equation:
v = vi + a*t
Where v is the final velocity, vi is the initial velocity (36.2 m/s), a is the acceleration (-9.8 m/s^2), and t is the time to strike the ground (3.696 s).
By substituting the known values into the equation, we can solve for v:
v = 36.2 + (-9.8)*3.696
v ≈ -0.247 m/s
Therefore, the ball will be moving at approximately -0.247 m/s just before it hits the ground.