Question:
An object of mass 3 kg falls from a height of 100 m onto sand and penetrates 2 m into the sand. How long did it take to enter the sand?
[A] 9s [B] 0.9 s [C] 0.09 s [D] 10 s
Answer:
Using the equation for motion:
v2 = u2 + 2as
where:
v = final velocity (0 m/s as it comes to rest)
u = initial velocity
a = acceleration (retardation in this case)
s = displacement (2 m)
We can rearrange the equation to solve for time (t):
u2 = 2as
u2 = 2 * (-490 m/s2)2 m
u2 = -1960
Taking the square root of both sides:
u ≈ √(-1960) ≈ ± 44.27 m/s
Since the object is falling, we take the negative value for the initial velocity:
u ≈ -44.27 m/s
Now, we can use the equation:
v = u - at
0 = -44.27 - (490 m/s2) * t
Simplifying the equation:
44.27 = 490t
t ≈ 44.27 / 490 ≈ 0.09 s
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