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 solve the problem, we will use the equations of motion and the given values of the problem. Let's solve each part step by step:
a) To find the maximum height of the professor, we need to determine when the professor's velocity becomes zero while moving upwards.
We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Given:
Initial velocity (u) = +10 ft/s (since it is upward)
Acceleration (a) = -32 ft/s^2 (negative since it acts downward due to gravity)
At the maximum height, the final velocity (v) will be zero. Thus, the equation becomes:
0 = (10 ft/s)^2 + 2(-32 ft/s^2)s
Simplifying,
0 = 100 ft^2/s^2 - 64 ft/s^2s
64 ft/s^2s = 100 ft^2/s^2
s = 100 ft^2/s^2 / 64 ft/s^2
s = 1.5625 ft
Therefore, the maximum height of the professor is 1.5625 feet.
b) To find the time before the professor hits the water (assuming "wimming with the fishes" means hitting the water surface), we need to calculate the time it takes for the professor to reach ground level.
We'll use the equation of motion: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity (u) at the maximum height = 0 ft/s
Acceleration (a) = -32 ft/s^2 (negative since it acts downward due to gravity)
Displacement (s) = 240 ft (height of the bridge)
240 ft = 0 ft/s * t + (1/2)(-32 ft/s^2)t^2
240 ft = (1/2)(-32 ft/s^2)t^2
480 ft = -32 ft/s^2t^2
t^2 = -480 ft / -32 ft/s^2
t^2 = 15 s^2
t = sqrt(15 s^2)
Therefore, the time it takes for the professor to hit the water is approximately sqrt(15) seconds.