A brick is dropped from a 78 m high scaffold. What is the velocity as the brick hits the pavement?

Question

It's a freefall question.

Answer 1

I'm fairly certain that you know a formula for distance falling under Earth's gravity as a function of time, and also a formula for velocity as a function of time.  Use them!  [g = 9.8 m/sec² in the MKS system]

 

s = -½gt²  ⇐    s = -78 meters...find out what t must be, then use t in

v = -gt   (this is the derivative w.r.t. time of the s formula)

Answer 2

Another way to solve this without first calculating the time is to use the kinematic formula for motion under constant acceleration:

 

v_f² - v_i² = 2ay

 

v_f = final speed (which is what you want to solve for)

v_i = initial speed = 0 in this problem (brick is dropped from rest)

a = constant acceleration = -g = -9.8 m/s² 

y = final position = -78 m (assuming the brick fell from an initial height of 0 m down to -78 m)

 

This simplifies to

 

v_f = √(2ay) = √[2(-9.8 m/s²)(-78 m)] = 39.10 m/s

 

This gives the same final speed as in Kenneth's solution, without having to first compute the time of flight.  When working problems like this, use the kinematic relation(s) if you do not need to know the time(s) involved.  If the problem also asks for time(s), then you have to solve it as Kenneth did.