Answer:
A) The total time it takes for the ball to reach the ground = 11.702 seconds
B) The time taken by the ball to reach the highest point = 3.8 seconds
Explanations:
To get the distance from the top of the building to the maximum height reached by the ball:

The initial speed, u = 38 m/s
The final speed at the maximum height, v = 0m/s
g = -10m/s² (Since the ball is thrown upwards)

The distance from the top of the building to the maximum height = 72.2 m
The time taken for the ball to reach the highest point from the top of the building will be calculated using the equation:
v = u + gt
0 = 38 + (-10)t
10t = 38
t = 38/10
t = 3.8 s
From the maximum height of the ball to the ground:
The height, H = 72.2 + 240
H = 312.2 m
The initial velocity, u = 0 m/s
g = 10 m/s²
Using the equation below:
![\begin{gathered} v^2=u^2+2gH \\ v^2=0^2\text{ + 2(10)}(312.2) \\ v^2\text{ = }6244 \\ v\text{ = }\sqrt[]{6244} \\ v\text{ = }79.02\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/cnhy51nb9ait5ldo4f5b1gsgwkfn00ojtm.png)
The time spent from the top of the building to the the ground will be calculated using the formula:
v = u + gt
79.02 = 0 + 10t
10t = 79.02
t = 79.02/10
t = 7.902 s
The total time it takes for the ball to reach the ground = 7.902 + 3.8
The total time it takes for the ball to reach the ground = 11.702 seconds
The time taken by the ball to reach the highest point = 3.8 seconds