Question

A stone weighing 3kg falls from the top of a tower 100m high and buries itself 2m in the ground. The time penetration is:

1. 0.09 sec
2. 0.9 sec
3. 2.1 sec
4. 1.3 sec

Answer 1

Given :
m = 3 kg

u = 0
h = 100 m
g = 9.8 m/s²

Using 3rd equation of motion :
v²-u² = 2gh

v²-0² = 2 (9.8) 100
v = 2√490

v = 14√10 m/s
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Now, during penetration :

u = 14√10 m/s
v = 0
S = 2 m
a = ?
t = ?

Using 3rd equation of motion;
v²-u² = 2aS

0²-(14√10)² = (2) a (2)
4a = -1960
a = -490 m/s²
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Now, using 1st equation is motion ;
v = u+at

0 = 14√10 + (-490)t
t = 14√10/490
t = 0.0903 sec
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Hence the correct alternative is
(1) 0.09 sec

Answer 2

The answer is option (1) 0.09 seconds

GIVEN

A stone weighing 3kg falls from the top of a tower 100m high and buries itself 2m in the ground.

TO FIND

The time of penetration.

SOLUTION

We can simply solve the above problem as follows;

Given,

Mass (m) = 3 kg

Initial velocity = 0 (free fall)

Height(H) = 100 meters

Acceleration (g) = 9.8 m/s²

Applying 3rd equation of motion

v² = u² + 2gH

v² = 2 × 9.8 × 100

v² = 1960

v = 14√10 m/s

Now,

During penetration

u = 14√10 m/s

v = 0

s = 2 meters

a = ?

t = ?

Applying 3rd equation of motion;

v² = u² + 2gh

-(14√10)² = 4a

4a = -1960

a = -1960/4 = -490 m/s²

Using first equation of motion

v = u + at

0 = 14√10 - 490t

-490t = -14√10

t = -14√10/-490

t = 0.09 seconds

Hence, The answer is option (1) 0.09 seconds

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