[niki]

I need help with these problems

1. a 1kg brick fall off a 3m wall onto a man's foot. How long does the brick take to fall. What is the Kenetic Energy of the brick relative to his foot.

[MathMaven53]

It's been years since I have studied Physics but I'll
work with the equations and you can substitute values into them.

Let s be distance brick falls,m the mass of the brick, g be acceleration due to gravity, t be time of fall, v be velocity of brick.

s = (1/2)g t^2

Solve for t

t = sqrt((2s)/g)

v = gt

= sqrt(2 g s)

Kinetic Energy is given by

(1/2) m v^2

= m g s

g is 32 ft/sec^2 in English system of measurement
but you are using metric. I forget what g is in the
metric system but it should be listed in your physics
textbook.

In summary: To find the time it takes for the brick to fall, use the equation

t = sqrt((2s)/g)

The brick's Kinetic Energy is given by m g s

[Polokid16]

Using SI (or english system)

"_" - subscript

E_p = mgh = (1kg)(9.8m/s^2)(3m) = 29.4J


For the time:

d = v_i * t +(0.5)at^2

since v_i or v_initial = 0m/s

d = (0.5)at^2

rearrange to solve for t:

sqrt( d / (0.5)g) = sqrt( 3m / (0.5)(9.8m/s^2)) = 0.6s

[itsme_vipsdude]

look as the body falls from rest therefore its initial kinetic energy is zero... therefore it must hve only the potential energy... potential energy
=mgh
=30*9.8*1
=294J
thus t the lowermost point kinetic energy is equal to the potential energy lost... tus k.e=294J

[Polokid16]

The wall is 3m... but yes E_k = E_p