Question

As you are cycling to classes one day, you pass a construction site on Green Street for a new building and stop to watch for a few minutes. A crane is lifting a batch of bricks on a pallet to an upper floor of the building. Suddenly, a brick falls off while the pallet is rising. You clock the time it takes the brick to hit the ground at 2.4 seconds. The crane, fortunately, has height markings, and you see the brick fall off the pallet at a height of 13 meters above the ground. A falling brick, as we all know, can be dangerous, and you wonder how fast the brick was going when it hit the ground. Since you are taking physics, you quickly calculate the answer.

Answer 1

The brick would hit the ground at 17.18m/s downward

v = -17.18 m/s

Step-by-step explanation:

Since the brick falls off while the pallet is rising it would have gained an instantaneous velocity (u) upward.

Using the equation of motion;

h = ut + 0.5at^2 ....1

For it to reach the ground h = 0, but there is an initial height of h0= 13 m

a = -g (acceleration due to gravity

Equation 1 becomes;

h = ut - 0.5gt^2 + h0 .....2

Substituting the values

t = 2.4s, g = 9.8m/s

Equation 2 becomes;

0 = 2.4u - (0.5×9.8×2.4^2) + 13

Solving for t,

2.4u = 15.224

u = 15.224/2.4

u = 6.34m/s

Hint : u = (0.5gt^2 - h0)/t

Applying another equation of motion.

v = u + at = u - gt

Final velocity of brick v = u - gt

v = 6.34 - 9.8(2.4)

v = -17.18 m/s

Answer 2

15.97 m/s

Step-by-step explanation:

If we ignore air resistance and apply the law of mechanical energy conservation, when the brick drops from 13m high above the ground, its potential energy is converted to kinetic energy:

`E_p = E_k`

`mgh = mv^2/2`

where m is the mass and h is the vertical distance traveled, v is the velocity at the ground

Let g = 9.81 m/s2. We can divide both sides by m

`gh = v^2/2`

`v^2 = 2gh = 2*9.81*13 = 255.06`

`v = √(255.06) = 15.97 m/s`