A cement block accidentally falls from rest from the ledge of a 51.6 -m-high building. When the block is 15.7 m above the ground, a man, 1.90 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Question

A cement block accidentally falls from rest from the ledge of a 51.6 -m-high building. When the block is 15.7 m above the ground, a man, 1.90 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Answer 1

To find the time the man has to get out of the way, we can use the equations of motion. First, let's determine the time it takes for the cement block to fall from the ledge to a height of 15.7 m above the ground:

We can use the equation of motion: h = ut + (1/2)gt^2, where
h is the height,
u is the initial velocity (0 m/s as the block is at rest),
g is the acceleration due to gravity (9.8 m/s^2),
t is the time.

Substituting the values:
15.7 = 0*t + (1/2)*(9.8)*t^2
15.7 = 4.9t^2

Rearranging the equation gives us a quadratic equation:
4.9t^2 - 15.7 = 0

Now, we can solve this quadratic equation to find the time it takes for the block to reach a height of 15.7 m:

Using the quadratic formula: t = (-b ± √(b^2 - (4ac)))/(2a)

In this case, a = 4.9, b = 0, and c = -15.7.

t = (-0 ± √(0 - (4*4.9*(-15.7)))) / (2*4.9)
t = (√(305.62)) / 9.8

Ignoring the negative solution, we have:
t ≈ √(31.223) / 9.8
t ≈ 5.001 seconds (approximately)

So it takes approximately 5.001 seconds for the cement block to fall from the ledge to a height of 15.7 m above the ground.

Next, we need to find the time it takes for the cement block to fall the remaining distance (51.6 - 15.7 = 35.9m) from the 15.7m height to the ground:

We can use the same equation: h = ut + (1/2)gt^2

Substituting the values:
35.9 = 0*t + (1/2)*(9.8)*t^2
35.9 = 4.9t^2

Again, solve the quadratic equation:
4.9t^2 - 35.9 = 0

Using the quadratic formula: t = (-b ± √(b^2 - (4ac)))/(2a)
a = 4.9, b = 0, and c = -35.9

t = (√(364.81)) / 9.8

Ignoring the negative solution:
t ≈ √(37.209) / 9.8
t ≈ 2.034 seconds (approximately)

Therefore, it takes approximately 2.034 seconds for the cement block to fall from the height of 15.7m to the ground.

To find the maximum time the man has to get out of the way, we can add the two calculated times:
Total time = 5.001 seconds + 2.034 seconds
Total time ≈ 7.035 seconds (approximately)

Hence, the man has approximately 7.035 seconds to get out of the way.

Answer 2

I think you mean a concrete block, not a cement block.

http://www.todayifoundout.com/index.php/2011/12/the-difference-between-cement-and-concrete/