(a) To find the coin's height above the river at t = 10 seconds, substitute t = 10 into the function H(t):

H(10) = 2116 - 16(10)^2
H(10) = 2116 - 16(100)
H(10) = 2116 - 1600
H(10) = 516 feet

So, the coin's height above the river at t = 10 seconds is 516 feet.

(b) The velocity of the coin at any given time t is the derivative of the height function, H(t). To find the velocity at t = 10 seconds, differentiate H(t) with respect to t:

H'(t) = -32t

Now substitute t = 10 into H'(t):

H'(10) = -32(10)
H'(10) = -320

So, the coin's velocity at t = 10 seconds is -320 ft/s. Note that the negative sign indicates that the coin is moving downward.

(c) To find how many seconds it will take for the coin to hit the water, we need to solve the equation H(t) = 0 (since the height above the water is 0 when the coin hits the water). Using the given function H(t):

2116 - 16t^2 = 0

Solving this quadratic equation, we can factor out a common factor of 4:

4(529 - 4t^2) = 0

Now we can set each factor equal to 0:

529 - 4t^2 = 0

Rearranging the equation:

4t^2 = 529

Dividing by 4:

t^2 = 529/4

Taking the square root of both sides:

t = √(529/4)

t = √(132.25)

t ≈ ±11.5

Since time cannot be negative in this context, the coin will hit the water approximately 11.5 seconds after it is dropped.

(d) The velocity of the coin as it strikes the water can be found by substituting t = 11.5 into H'(t):

H'(11.5) = -32(11.5)
H'(11.5) = -368

So, the coin's velocity as it strikes the water is -368 ft/s. Again, the negative sign indicates that the coin is moving downward.

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