Help me slove this

Question

 There is an old urban myth that states if you drop a penny off of the Empire State Building and it hits someone walking on the street below, it could kill them. If the Empire State Building is 381 m tall, and a penny happened to fall off the top, 

a. How long would it take to strike someone in the head that is 1.8 m tall? b. How fast would the penny be travelling if it missed the passer-by and hit the ground? 

c. How long would it take the penny to fall the first 100 m? 

d. How fast would the penny be travelling at time t = 4.5 s? 

Answer 1

Use the kinematic equations

1. v(f) = v(o) + at
2. v(f)^2 = v(o)^2 + 2ax
3. x = v(o)t + 1/2(a)(t)^2

a. How long would it take to strike someone in the head that is 1.8 m tall?

If the total distance to the ground is 381m and somebody is 1.8m higher than the ground, the total distance we need to use is 381m - 1.8m = 379.2m.

displacement = 379.2 m

initial velocity = 0 m/s

acceleration = -9.8 m/s^2

time = ?

Plug into equation 3: 379.2m = (0m/s)(t) + 1/2 (-9.8m/s/s)(t)^2

solving for t gives us : t= 9.079 seconds

b. How fast would the penny be travelling if it missed the passer-by and hit the ground? 

in this case:

displacement = 381m

velocity initial = 0 m/s

velocity final = ? m/s

acceleration = -9.8 m/s/s

Plug into equation 2: v(f)^2 = 0 + 2(-9.8 m/s/s)(381m)

solving for v(f) gives us: v(f) = -86.41 m/s

c. How long would it take the penny to fall the first 100 m? 

x = 100 m

v(o) = 0 m/s

a = -9.8 m/s/s

t= ?

Plug into equation 3: x = v(o)t + 1/2(a)(t)^2

100m = 0 + 1/2(-9.8m/s/s)(t)^2

(100m) / (4.6m/s/s) = 21.739

21.739 = t^2

t= 4.66 seconds

d. How fast would the penny be travelling at time t = 4.5 s? 

t = 4.5 s

v(f) = ? m/s

v(o) = 0 m/s

a = -9.8 m/s/s

Plug into equation 1: v(f) = v(o) + at

v(f) = 0 m/s + (-9.8m/s/s)(4.5 s)

v(f) = -44.1 m/s