Question

A brick falls off the top of a wall under construction and drops into a bed of sand 14.5m below. It makes a dent in the sand 185mm deep. What is
i) The speed of the brick just before it hits the sand
ii) Its deceleration in the sand
iii) Time it takes to fall

Answer 1

Explanation:

the velocity of the brick as it touches the sand will be

v = [2gh]1/2

[from v2 - u2 = 2as ; u = 0, a = g and s = h]

here,

h = 14.5 m

so,

v = [2×10×14.5]1/2

or

v = 17 m/s

and

the time taken by the brick to reach the sand

t = v / h = 17 / 14.5

thus,

t = 1.17 s

now,

the brick travels through the sand and finally comes to rest. In this case

the deceleration will be

a = (v2 - u2) / 2s

[from v2 - u2 = 2as]

here,

v = 0

u = 17 m/s

s = 185mm = 0.185m

so,

a = (0 - 172) / 0.185

thus,

a = -1562.162 m/s2