An iron ball is dropped from rest, it falls 250 m before hitting the ground. what is the velocity just before it hits the ground?

Question

i need help doing my physical science work. this is an online course, therefore it can be difficult to receive help with the work. 

Answer 1

Height as a function of time:

h(t) = (-1/2)gt² + v₀t + h₀

g = acceleration of gravity = 9.8 m/s²

v₀ = initial velocity = 0

t = time

h₀ = initial height = 250

 

h(t) = -4.9t² + 250 

 

We want the time when h(t) = 0

-4.9t² + 250 = 0

-4.9t² = -250

t² = 250/4.9      note that the negatives cancel out

t = √(250/4.9)

t = 7.143 seconds

 

The dropped object gains 9.8 meters per second² for 7.143 seconds:

Final velocity = (9.8 m/s²)(7.143 s) = 70 m/s final velocity

 

Answer 2

Knowing its acceleration is constant (g = 9.8 m/s²), use the kinematic relation

 

v_f² - v_i² = 2ay = 2gy

 

v_i = 0  [started from rest]

y = vertical fall distance = 250 m

v_f = final speed = ?

 

v_f = √(2gy) = √[2(9.8)(250)] m/s = ? m/s