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 kinematic equation that relates initial velocity (v0), final velocity (vf), acceleration (a), and displacement (d):
vf^2 = v0^2 + 2ad
In this case, the initial velocity is 10ft/s, the acceleration due to gravity is -32ft/s^2, and the displacement is the maximum height h. The final velocity when reaching the maximum height is 0ft/s since it stops momentarily before falling back down.
Therefore, the equation becomes:
0^2 = (10ft/s)^2 + 2(-32ft/s^2)h
Simplifying the equation:
0 = 100ft^2/s^2 - 64ft/s^2 * h
Solving for h:
64ft/s^2 * h = 100ft^2/s^2
h = 100ft^2/s^2 / 64ft/s^2
h = 1.5625ft^2
Therefore, the maximum height of the professor is approximately 1.5625ft.
To find the time before the professor hits the water, we can use another kinematic equation that relates initial velocity (v0), final velocity (vf), acceleration (a), and time (t):
vf = v0 + at
In this case, the initial velocity is 10ft/s, the acceleration due to gravity is -32ft/s^2, and the final velocity when hitting the water is unknown (we'll call it vw). We need to solve for time (t).
The equation becomes:
vw = 10ft/s + (-32ft/s^2) * t
Since vw is when the professor hits the water, vw = 0ft/s.
0ft/s = 10ft/s + (-32ft/s^2) * t
Simplifying the equation:
-10ft/s = -32ft/s^2 * t
Solving for t:
t = -10ft/s / -32ft/s^2
t = 0.3125s
Therefore, the professor will hit the water approximately 0.3125 seconds after being thrown off the bridge.