while at the shooting range, jonathan shoots a bullet traveling horizontally with speed of 35 m/s. the bullet hits a board, perpendicular to the surface, passes through it and emerges on the other side with speed of 21 m/s. if the board is 4.00cm thick, how long does the bullet take to pass through the board?

a = (V^2-Vo^2)/2d

a = (21^2-35^2/0.08 = -9800 m/s^2.

V = Vo + a*t = 21 m/s.
35 + -9800t = 21
t = 0.001429 s. = 1.429 ms.

To find the time it takes for the bullet to pass through the board, we can use the equation:

time = distance / velocity

First, let's calculate the distance the bullet travels through the board. Since the board is perpendicular to the bullet's motion, the distance it travels through the board will be the same as the thickness of the board.

Given:
Bullet's initial velocity (u) = 35 m/s
Bullet's final velocity (v) = 21 m/s
Board thickness (d) = 4.00 cm = 0.04 m

The change in velocity (Δv) is given by:
Δv = v - u = 21 m/s - 35 m/s = -14 m/s

We know that acceleration (a) is given by:
a = Δv / t

Since there is no force acting on the bullet in the horizontal direction, its acceleration is zero. Hence, Δv = 0 and the bullet's velocity remains constant while passing through the board.

Now, substituting the values into our equation for acceleration, we have:
0 = 0 / t

Since anything divided by zero is undefined, it means the bullet passes through the board instantaneously.

Therefore, the time taken for the bullet to pass through the 4.00 cm thick board is effectively zero seconds.