Brick Falling Off a Rising Pallet

[alandry06]

Homework Statement

A crane is in the process of lifting a pallet of bricks to the upper floor of a building. Suddenly a brick slides and falls off the rising pallet. You clock the time it takes for the brick to hit the ground at 2.5 seconds.

The crane has height marking and you see that the brick fell off the pallet from a height of 22 meters. How fast is the brick going when it hits the ground?

Homework Equations

y = y_o + V_ot + (1/2)at²

V = V_o + at

The Attempt at a Solution

So we are given that:

t = 2.5 s
y = 22 m
y_o = 0
a = g = 9.80 m/s²
V_o = ?

Find: V = ?

I understand that I need to use the kinematic equations to solve this. What I am confused by is that the words 'lifting' and 'rising' are emphasized in the problem statement.

Does this imply that V_o is NOT equal to 0 ?

Is so, I am confused as to how to find it?

Any advice is appreciated

 

[rl.bhat]

Brick slides off from the rising pallet with some velocity. Before that brick and pallet have the same velocity. So when the brick slips form the rising pallet, what will be its initial velocity and in which direction?

 

[alandry06]

It will be up; but I do not know its value. Or is it just V_o?

Can I solve the 2 equations in 2 unknowns now? Is V_o the same for both of the equations?

Thank you!

 

[rl.bhat]

Velocity vo is the same as the velocity of the pallet.
Use the first relevant equation.
In that y = 0, yo = 22 m. t is given. Find vo. Take care of the directions of the various terms.

 

[alandry06]

Can I use y₀= 0 and y = 22 if I choose positive as going downwards?

So,

y = V_ot + at²

V_o = (y - at²) / t

V_o = (22 - 9.80*2.5²) / 2.5

V_o = - 15.7 m/s

So V_o is negative because it is going upwards and I chose downwards as positive.

Now I use this in the second equation. Again choosing down as positive

V = V_o + at

V = (-15.7) + 9.8 * 2.5

V = 8.8 m/s going downwards.

Does this look correct?

 

[rl.bhat]

y = V_ot + at²
It should be
y = V_ot + 1/2*at²

 

[alandry06]

Oops.* So it should be"y = V_ot + (1/2)at²

V_o = (y - (1/2)at²) / t

V_o = (22 - 0.5*9.80*2.5²) / 2.5

V_o = - 3.45 m/s

So V_o is negative because it is going upwards and I chose downwards as positive.

Now I use this in the second equation.* Again choosing down as positive

V = V_o + at

V = (-3.45) + 9.8 * 2.5

V = 21.1 m/s going downwards.

Is this correct?* It seems like a better answer in that the first one (with the error) was too slow.

 

[rl.bhat]

Yes. It is correct.