a stone weighing 3kg falls from the top of a tower 100m high and buries itself 2m deep in the sand .What is the time of penetration ?
Answers
Answered by
7
Answer:
ANSWER
The velocity of the ball before reaching the ground is given by
v
2
=u
2
+2as
v
2
=2×9.8×100
v=14
10
Distance penetrated before coming to rest =2m
If the retardation experienced while penetrating is a
v
2
=u
2
−2as
v=0
u
2
=2as
a=490m/s
2
retardation
v=u−at
t=
490
14
10
=0.09s answer is =0.09s
Answered by
0
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
___________________________
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
------------------------------------------
Now, using 1st equation is motion ;
v = u+at
0 = 14√10 + (-490)t
t = 14√10/490
t = 0.0903 sec
===========================
Hence the correct alternative is
(1) 0.09 sec
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