The velocity of the ball when it was caught is 12.52 m/s.
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find the velocity of the ball when it was caught.
The given parameters;
maximum height above the ground reached by the ball, H = 38 m
height above the ground where the ball was caught, h = 30 m
The height traveled by the ball when it was caught is calculated as follows;
y = H - h
y = 38 - 30 = 8 m
The velocity of the ball when it was caught is calculated as;

Thus, the velocity of the ball when it was caught is 12.52 m/s.
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The volume of the cylindrical can is given by:
V = πr²h
V = volume, r = base radius, h = height
Differentiate both sides of the equation with respect to time t. The radius r doesn't change over time, so we treat it as a constant:
dV/dt = πr²(dh/dt)
Given values:
dV/dt = -527in³/min
r = 8in
Plug in and solve for dh/dt:
-527 = π(8)²(dh/dt)
dh/dt = -2.62in/min
The height of the water is decreasing at a rate of 2.62in/min