Superman lived on Krypton where the acceleration due to gravity is 23.6 m/s2. If a woman falls from a building that is 88.3 m tall, how long does he have to save her before she hits the ground?
Givens:
Unknown:
Substitute:
Solve:
88.3 = (1/2) g t^2
88.3 = .5 * 23.6 * t^2
t = 2.74 seconds
To solve this problem, we can use the kinematic equation for displacement:
d = v0t + (1/2)at^2
Where:
d = displacement (in this case, equal to the height of the building)
v0 = initial velocity (in this case, assumed to be 0 since the woman is falling)
a = acceleration due to gravity (23.6 m/s^2)
t = time
Givens:
Height of the building = 88.3 m
Acceleration due to gravity on Krypton = 23.6 m/s^2
Unknown:
Time it takes for the woman to fall (t)
Substitute the given values:
d = 88.3 m
a = 23.6 m/s^2
v0 = 0 (since the woman is falling, she has no initial velocity)
Now, let's solve for t:
88.3 m = 0(t) + (1/2)(23.6 m/s^2)t^2
Simplifying the equation:
44.15 t^2 = 88.3 m
Divide both sides of the equation by 44.15:
t^2 = 2
Take the square root of both sides to solve for t:
t ≈ √2 seconds
Therefore, Superman has approximately √2 seconds to save the woman before she hits the ground.