A diver is on the 10 m platform, preparing to perform a dive. The diver’s height above the water, in metres, at time t can be modelled using the equation

h(t)= 10+ 2t - 4.9t^2.

a) Determine when the diver will enter the water.
b) Estimate the rate at which the diver’s height above the water is
changing as the diver enters the water.

a) Ah, the eternal question: When will the diver enter the water? Well, we can find that out by setting h(t) equal to zero. So let's solve the equation 10 + 2t - 4.9t^2 = 0. Or we can just wait and see if they belly flop or gracefully swan dive. The suspense is killing me!

b) Now, the rate at which the diver's height above the water is changing as they enter the water can be found by taking the derivative of h(t) with respect to time. But let's be honest, it's going to be a pretty rapid change. From high above to a splash in seconds! Better call it a "plunge rate" rather than a "rate of change".

a) To determine when the diver will enter the water, we need to find the value of t when h(t) equals 0. In other words, we need to solve the equation:

10 + 2t - 4.9t^2 = 0.

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac))/(2a),

where a = -4.9, b = 2, and c = 10. Plugging in these values, we have:

t = (-2 ± √(2^2 - 4(-4.9)(10)))/(2(-4.9)).

Simplifying further, we have:

t = (-2 ± √(4 + 196))/(-9.8).

Taking the positive value, we get:

t = (-2 + √200)/(-9.8).

Now, calculating the approximate value using a calculator, we find:

t ≈ 2.018.

Therefore, the diver will enter the water after approximately 2.018 seconds.

b) To estimate the rate at which the diver's height above the water is changing as the diver enters the water, we need to find the derivative of h(t) with respect to t. Taking the derivative, we have:

h'(t) = 2 - 9.8t.

Now, plugging in the value of t when the diver enters the water (which we found in part a), we have:

h'(2.018) = 2 - 9.8(2.018).

Simplifying further, we get:

h'(2.018) ≈ 2 - 19.9616 ≈ -17.9616.

Therefore, the rate at which the diver's height above the water is changing as the diver enters the water is approximately -17.9616 m/s.

To determine when the diver will enter the water, we need to find the time at which the height above the water, h(t), equals zero.

a) So, to solve for t, we need to set h(t) equal to zero and solve the resulting equation:

0 = 10 + 2t - 4.9t^2

This is a quadratic equation, so we can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -4.9, b = 2, and c = 10.

Plugging in the values, we have:

t = ([-2] ± √([2]^2 - 4[-4.9][10])) / (2[-4.9])

Simplifying this expression further:

t = (-2 ± √(4 + 196)) / (-9.8)

t = (-2 ± √200) / (-9.8)

Now, evaluating this expression:

t ≈ (-2 ± 14.14) / (-9.8)

We get two possible solutions for t: t ≈ 1.41 seconds and t ≈ -1.63 seconds. Since time cannot be negative, we discard the negative solution. Therefore, the diver will enter the water approximately 1.41 seconds after starting the dive.

b) To estimate the rate at which the diver's height above the water is changing as they enter the water, we need to find the derivative of h(t) with respect to time (t). The derivative gives us the rate of change of the function.

Differentiating h(t) = 10 + 2t - 4.9t^2 with respect to t:

h'(t) = 2 - 9.8t

Now, we substitute t = 1.41 (the time when the diver enters the water) into the expression we just derived:

h'(1.41) = 2 - 9.8(1.41)

Evaluating this expression:

h'(1.41) ≈ 2 - 13.78

h'(1.41) ≈ -11.78

Therefore, the estimated rate at which the diver's height above the water is changing as they enter the water is approximately -11.78 meters per second. The negative sign indicates that the height is decreasing as the diver enters the water.

10 + 2t - 4.9t² = 0

t = 1.647 s

b) take the derivative and plug in 1.647 s

dh/dt = 2 - 9.8 t
dh/dt = 2 - 9.8 (1.647 ) = -14.14 m/s