A brick is dropped from the top of a building that is 45m high. A man on the sid

Question

A brick is dropped from the top of a building that is 45m high. A man on the sidewalk below the building is moving with constant velocity, and is 12m from the point of impact when the brick is dropped. What is the speed of the man if he is to be hit with the brick?

I did....

x = x0 + v0x(t) + (.5)ax(t^2)

So 0 = -45 + 0(t) + (.5)(9.8)(t^2)

0 = -45 + (.49)(t^2)

t^2 = 9.1836

t = 3.03 for the brick to hit the sidewalk


Then

x - x0 = ((v0x+vx)/2)t

-12 - 0 = ((v0x +vx)/2)t

-12 = vx(t)

vx = 3.96 meters per second


Is this correct?

Explanation / Answer

Part 1 is right. Though, to make things clearer, I'm going to re-do it with two changes: use y instead of x for clarity, and make up the positive y-direction.

y(t) = 45 + v(0)*t + (1/2)(-9.8)t2

y(t) = 45 - 4.9t2

We want to know when it hits the ground, i.e., what is t when y(t)=0.

0 = 45 - 4.9t2

t2 = 45/4.9 = 9.18

t = 3.03 s. Good! We agree.

Now, the man is walking at constant velocity in the x-direction starting from 12 meters away. Let's make the point of impact where x and y both equal 0. So x(0) = -12

x(t) = x(0) + v*t

0 = -12 + v(3.03)

12 = v(3.03)

v = 12/3.03 = 3.96 m/s

The man was walking 3.96 m/s before being hit by a brick.

So, congrats! Your answer agrees with mine. yay.

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