In the movie Superman, Lois Lane falls from a building and is caught by the diving superhero. Assuming that Lois, with a mass of 45 kg, is falling at a terminal velocity of 59 m/s, how much force does Superman exert on her if it takes 0.4 s to slow her to a stop?



If Lois can withstand a maximum acceleration of 7 g's, what minimum time should it take Superman to stop her after he begins to slow her down?

force*time= mass*change in velocity

solve for force.

F*time=mass*changevelocity
7g*m*time=mass*change velocity
solve for time

Ei = Ef

S-m*g = (m*V)/(delta t)
solve for S

Step 1: Calculate the initial velocity of Lois Lane using the formula: v = gt, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time taken (0.4 s).

v = gt
v = 9.8 m/s^2 * 0.4 s
v = 3.92 m/s

Step 2: Calculate the deceleration required to stop Lois Lane using the formula: a = (vf - vi) / t, where vf is the final velocity (0 m/s), vi is the initial velocity (3.92 m/s), and t is the time taken (0.4 s).
a = (0 m/s - 3.92 m/s) / 0.4 s
a = -3.92 m/s / 0.4 s
a = -9.8 m/s^2 (Note: The negative sign indicates deceleration)

Step 3: Calculate the force exerted by Superman on Lois Lane using the formula: F = ma, where m is the mass of Lois Lane (45 kg) and a is the acceleration (-9.8 m/s^2).
F = ma
F = 45 kg * -9.8 m/s^2
F = -441 N

Therefore, Superman exerts a force of 441 Newtons on Lois Lane to slow her to a stop in 0.4 seconds.

To find the minimum time required to stop Lois Lane within her maximum acceleration tolerance:

Step 4: Calculate the maximum acceleration in terms of m/s^2 using the formula: a = g * 7, where g is the acceleration due to gravity (9.8 m/s^2) and 7 is the acceleration tolerance (7 g's).
a = g * 7
a = 9.8 m/s^2 * 7
a = 68.6 m/s^2

Step 5: Calculate the deceleration time using the formula: t = (vf - vi) / a, where vf is the final velocity (0 m/s), vi is the initial velocity (3.92 m/s), and a is the maximum acceleration (68.6 m/s^2).
t = (0 m/s - 3.92 m/s) / 68.6 m/s^2
t = -3.92 m/s / 68.6 m/s^2
t = -0.0572 s (Note: The negative sign indicates deceleration time)

Since time cannot be negative, the minimum time required to stop Lois Lane is 0.0572 seconds (or approximately 0.06 seconds).

To calculate the force exerted by Superman on Lois Lane, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the acceleration can be calculated by using the formula a = (final velocity - initial velocity) / time.

To get the initial velocity, we know that Lois is falling at a terminal velocity of 59 m/s. Since terminal velocity is the constant speed reached when the force of gravity is balanced by the force of air resistance, we can assume that Lois initially falls with a velocity close to 59 m/s.

The final velocity is zero since Superman stops her completely. The time taken to stop her is given as 0.4 s.

So, let's calculate the acceleration first:
a = (final velocity - initial velocity) / time
a = (0 - 59) / 0.4
a = -147.5 m/s²

Now, we can calculate the force exerted by Superman:
F = mass * acceleration
F = 45 kg * -147.5 m/s²
F = -6637.5 N (negative sign indicates the direction opposite to the motion)

Therefore, Superman exerts a force of approximately 6637.5 N on Lois Lane to stop her.

Moving on to the second part of the question, we know that Lois can withstand a maximum acceleration of 7 g's. To find the minimum time needed for Superman to stop her, we can use the formula:

a = (final velocity - initial velocity) / time

Rearranging the formula, we have:

time = (final velocity - initial velocity) / acceleration

Substituting the values:

time = (0 - 59) / (7 * 9.8) (1 g = 9.8 m/s²)
time = -59 / 68.6
time ≈ -0.86 s

Since time cannot be negative, we take the positive value. Therefore, the minimum time it should take Superman to stop Lois after he begins to slow her down is approximately 0.86 seconds.