A cement block accidentally falls from rest from the ledge of a 55.9-m-high building. When the block is 11.2 m above the ground, a man, 1.70 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?

0.3 seconds approximately

Explanation:

Given that the height of the building is 55.9 m and cement block accidentally falls from rest from the edge of a 55.9-m-high building. The initial velocity of the block will be equal to zero. The final velocity will be achieved by using the formula:

V^2 = U^2 + 2gH

Where g = 9.8 m/s^2

H = 55.9m

V^2 = 0 + 2 × 9.8 × 55.9

V^2 = 1095.64

V = sqrt(1095.64)

V = 33.1 m/s

The velocity When the block is 11.2 m above the ground will be 33.1 m/s and the height h = 11.2 - 1.7 = 9.5 m

The time of escape can be calculated by using the formula;

h = Ut + 1/2gt^2

9.5 = 33.1t + 1/2 × 9.8 × t^2

9.5 = 33.1t + 4.9t^2

4.9t^2 + 33.1t - 9.5

By using quadratic formula to calculate the time, the negative value is ignored and the positive values will be the time the man has to get out of the way.

Please find the attached file for the remaining solution.

the answer would be a

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