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 reached by the professor, we need to determine when the professor's vertical velocity becomes zero. This occurs because at the maximum height, the velocity changes direction from upward to downward.
Let's use the kinematic equation for vertical motion:
vf = vi + at
Where:
- vf is the final velocity (zero in this case),
- vi is the initial velocity (10 ft/s upward),
- a is the acceleration (-32 ft/s^2 due to gravity), and
- t is the time.
a) Finding the maximum height:
We want to solve for t when vf = 0:
0 = 10 ft/s - 32 ft/s^2 * t
Rearranging the equation, we get:
32 ft/s^2 * t = 10 ft/s
t = 10 ft/s / 32 ft/s^2
t ≈ 0.3125 s
Now, we can find the maximum height using the equation for vertical displacement:
Δy = vit + 0.5at^2
Where:
- Δy is the vertical displacement (maximum height),
- vi is the initial velocity (10 ft/s upward),
- a is the acceleration (-32 ft/s^2 due to gravity), and
- t is the time.
Plugging in the values:
Δy = (10 ft/s)(0.3125 s) + 0.5(-32 ft/s^2)(0.3125 s)^2
Δy ≈ 1.5625 ft
Therefore, the maximum height reached by the professor is approximately 1.5625 feet above the initial position.
b) Finding the time before the professor hits the water:
We can use the equation for vertical displacement to find the time when the professor reaches the water's surface. At this point, the displacement would be -240 ft (negative due to the downward direction).
-240 ft = (10 ft/s)t + 0.5(-32 ft/s^2)t^2
Rearranging the equation:
0.5(-32 ft/s^2)t^2 + (10 ft/s)t - 240 ft = 0
Now, we can solve this quadratic equation to find the time (t) when it can hit the water's surface.