A 5 gm bulet initial velocity is 300ms-1
and after penetrating a 5cm plank then the bulet stopped . So how much time is needed to penetrate the plank..?
V^2 = Vo^2 + 2a*d.
0 = 300^2 + 2a*0.05, a = -900,000 m/s^2.
V = Vo + a*t.
0 = 300-900,000t, t = ?.
To find the time needed to penetrate the plank, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the bullet stops)
u = initial velocity (300 m/s)
a = acceleration (unknown)
s = distance (5 cm = 0.05 m)
First, let's rearrange the equation to solve for acceleration:
v^2 = u^2 + 2as
0^2 = (300 m/s)^2 + 2a(0.05 m)
0 = 90000 m^2/s^2 + 0.1a
Rearranging again, we have:
0.1a = -90000 m^2/s^2
a = -90000 m^2/s^2 / 0.1
a = -900000 m^2/s^2
Since the acceleration is negative, the bullet is decelerating.
Now, we can use another equation to find the time taken to reach the final velocity of 0:
v = u + at
0 = 300 m/s + (-900000 m^2/s^2)t
Simplifying this equation gives:
-300 m/s = -900000 m^2/s^2 t
Dividing both sides by -900000 m^2/s^2:
t = -300 m/s / -900000 m^2/s^2
t = 1/3000 s
Therefore, it takes t = 1/3000 s (or approximately 0.00033 seconds) for the bullet to penetrate the plank and come to a stop.