A cement block accidentally falls from rest from the ledge of a 51.7-m-high building. When the block is 14.9 m above the ground, a man, 2.05 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?
free fall from 51.7 to 14.9 meters - how fast is it going?
easy way is energy
(1/2) m v^2 = m g (51.7-14.9)
v^2 = 2 * 9.81 * 36.8
v = 26.7 m/s initial speed down at 14.9 height
how long to fall another (14.9 -2.05)= 12.85 meters?
increase in Ke = m g h
(1/2) m v^2 = (1/2)m(26.7)^2 + m g (12.85)
v^2 = 2* 26.7^2 + 2*9.81(12.85)
v^2 = 1677
v = 41 m/s when it hopefully misses
average speed between sighting and missing = .5(41 + 26.7) = 33.85 m/s
time to fall 12.85 m = 12.85/33.85 = .38 seconds (forget it)
Your formula is wrong, i tried it twice already.
To find the maximum time the man has to get out of the way, we need to calculate the time it takes for the block to fall from its initial position to a height of 14.9 m above the ground. We can use the equation of motion for free fall:
h = ut + (1/2)gt²
Where:
h = height
u = initial velocity (which is zero since the block is dropped from rest)
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time
Since the block is dropped from rest, the initial velocity is zero. The acceleration due to gravity is always acting downwards, so we can consider it as a negative value.
Let's solve the equation for the time it takes to reach a height of 14.9 m:
14.9 = 0t + (1/2)(-9.8)t²
14.9 = -4.9t²
Rearranging the equation, we get:
t² = -14.9 / -4.9
t² = 3.04
Taking the square root of both sides:
t ≈ √3.04
t ≈ 1.74 seconds
Therefore, it takes approximately 1.74 seconds for the block to fall to a height of 14.9 m above the ground.
Since the man noticed the block when it was at the same height as him (2.05 m), we can calculate the total time the man has to get out of the way. Subtracting the time it took to reach 14.9 m from the total time:
Total time = 1.74 seconds
Therefore, the man has approximately 1.74 seconds to get out of the way.