Suppose a coin is dropped from the top of the Empire State building in New York, which is 1,454 feet tall. The position function for free-falling objects is:
s(t) = −16t^2 + v_0t + s_0, where v_0 is the initial velocity and s_0 is the initial position.
Find the velocity of the coin just before it hits the ground.
When a coin is dropped:
t = 0
the initial velocity v₀ = 0
Substitute v₀ = 0 and s₀ = 1454 ft into the position function:
s(t) = - 16 t² + v₀ t + s₀
which becomes:
s(t) = - 16 t² + 1454
The velocity is the first derivative of s(t):
v(t) = ds / dt = s'(t) = - 32 t
When the coin reaches the ground level then s = 0
Solve:
s = 0
- 16 t² + 1454 = 0
Subtract 1454 to both sides
- 16 t² = - 1454
Multiply both sides by - 1
16 t² = 1454
t² = 1454 / 16
t = ± √1454 / √16
t = ± 38.131352 / 4
t = ± 9.532838
Time can not be negative so:
t = 9.532838 sec
v(t) = ds / dt = - 32 t
v(9.532838) = - 32 t = - 32 ∙ 9.532838 = - 305.050816 ft / s ≈ - 305 ft / s
The negative sign of velocity is obtained because the positive direction for s(t) is directed upwards and the negative one is directed downwards. So you need to take the absolute value of the speed.
v = | - 305 ft / s | = 305 ft / s