a math professor is thrown off the matthews bridges, height 240ft, by a disgruntled calculus class with an intitial upward velocity of 10ft/s

a) assumign the acceleration dude to gravity is -32ft/s^2,find the macimum height of the professor.
b) find the time before the professor is wimming with the fishes

To find the maximum height of the professor, we can use the equation of motion for an object under constant acceleration. The equation is given by:

h = h0 + v0t + (1/2)at^2

where:
h is the height at time t
h0 is the initial height (240 ft)
v0 is the initial velocity (10 ft/s)
a is the acceleration due to gravity (-32 ft/s²)
t is the time

a) To find the maximum height (h), we need to find the time (t) when the velocity becomes zero. At the maximum height, the velocity will be zero before it starts falling back down.

Using the equation for velocity under constant acceleration:

v = v0 + at

At the maximum height, v = 0, so we can solve for the time:

0 = 10 - 32t

Solving for t:

32t = 10
t = 10/32
t = 0.3125 seconds

Now, substitute the value of t into the equation for height:

h = h0 + v0t + (1/2)at^2
h = 240 + 10 * 0.3125 + (1/2)(-32)(0.3125)^2
h = 240 + 3.125 - 1
h ≈ 242.125 ft

Therefore, the maximum height reached by the professor is approximately 242.125 ft.

b) To find the time before the professor hits the water (wimming with the fishes), we can determine the time it takes for the professor to reach a height of 0 ft from the initial height of 240 ft.

Using the equation for height:

h = h0 + v0t + (1/2)at^2

Setting h = 0, h0 = 240 ft, v0 = 10 ft/s, and a = -32 ft/s², we can solve for t:

0 = 240 + 10t + (1/2)(-32)t^2

Simplifying the equation:

16t^2 + 10t - 240 = 0

We can solve this equation by factoring or using the quadratic formula. Factoring might be challenging, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For the quadratic equation 16t^2 + 10t - 240 = 0:

a = 16, b = 10, c = -240

Substituting the values into the quadratic formula:

t = (-10 ± √(10^2 - 4 * 16 * -240)) / (2 * 16)
t = (-10 ± √(100 + 15360)) / 32
t ≈ (-10 ± √15460) / 32

Calculating the values:

t ≈ (-10 ± 124.3)/32

Using both positive and negative values for the square root:

t1 ≈ (-10 + 124.3) / 32 ≈ 0.397 seconds
t2 ≈ (-10 - 124.3) / 32 ≈ -4.076 seconds

The negative value (-4.076 seconds) is discarded since the time cannot be negative in this context. Thus, the professor will hit the water approximately 0.397 seconds after being thrown off the bridge.