A 4.4 g bullet is fired into a block of wood with a mass of 21.3 g. The wood block is initially at rest on a 1.3 m tall post. After the collision, the wood block and bullet land 2.4 m from the base of the post. Find the initial speed of the bullet.

m1 =4.4•10^-3 kg, m2 = 21.3•10^-3 kg, h =1.3 m, L= 2.4 m., v1 =?

m1 •v1 = (m1+m2) •v,
v= m1 •v1 /(m1 + m2) .

h =g•t²/2 => t = sqrt(2•h/g)
L= v•t = v• sqrt(2•h/g),
v =L/sqrt(2•h/g),

m1 •v1 /(m1 + m2) = L/sqrt(2•h/g).

v1= L•(m1 + m2)/m1• sqrt(2•h/g) =
= 2.4•(4.4•10^-3+21.3•10^-3)/( 4.4•10^-3)•sqrt(2•1.3/9.8) =
= 2.4•2.57•10^-2/(4.4•10^-3)•0.515 =27.22 m/s.

Elena is right, but MAKE SURE that in the final step when calculating you put the (m1 times sqrt(2 times h/g)) all in parentheses.

To find the initial speed of the bullet, we can use the principle of conservation of momentum. The equation for momentum is:

Momentum = mass * velocity

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the wood block is initially at rest, its initial momentum is zero.

Let's denote the initial velocity of the bullet by "v". After the collision, both the bullet and the wood block move together. Let's denote their final velocity by "vf".

To solve this problem, we need to express the initial momentum and final momentum in terms of mass and velocity.

The initial momentum (before the collision) is given by the product of the bullet's mass and its initial velocity:

Initial momentum = (mass of bullet) * (initial velocity of bullet) = (4.4 g) * (v)

The final momentum (after the collision) is given by the sum of the bullet's and wood block's masses, multiplied by their final velocity:

Final momentum = (mass of bullet + mass of wood block) * (final velocity of bullet and wood block) = (4.4 g + 21.3 g) * (vf)

According to the principle of conservation of momentum, the initial momentum is equal to the final momentum:

(4.4 g) * (v) = (4.4 g + 21.3 g) * (vf)

Now, let's substitute the given values into the equation:

(4.4 g) * (v) = (4.4 g + 21.3 g) * (vf)

To make things easier, let's convert grams to kilograms (1 g = 0.001 kg):

(0.0044 kg) * (v) = (0.0044 kg + 0.0213 kg) * (vf)

Now, let's simplify the equation:

(0.0044 kg) * (v) = (0.0257 kg) * (vf)

Divide both sides of the equation by (0.0044 kg):

v = (0.0257 kg) * (vf) / (0.0044 kg)

Now, we know that the bullet and wood block land 2.4 m from the base of the post. We can use this information to find the final velocity (vf) of the bullet and wood block.

The equation to find the final velocity (vf) is:

vf^2 = vi^2 + 2 * a * d

where vi is the initial velocity, a is the acceleration, and d is the displacement.

In this case, since the bullet and wood block land vertically, the acceleration (a) is equal to the acceleration due to gravity (g) which is approximately 9.8 m/s^2. The displacement (d) is given as 2.4 m.

Now, let's solve the equation for vf:

vf^2 = 0^2 + 2 * (9.8 m/s^2) * (2.4 m)

vf^2 ≈ 45.12 m^2/s^2

Take the square root of both sides to find vf:

vf ≈ √(45.12 m^2/s^2) ≈ 6.71 m/s

Now, substitute this value of vf into the equation we derived earlier to find the initial velocity (v) of the bullet:

v = (0.0257 kg) * (6.71 m/s) / (0.0044 kg)

v ≈ 39.15 m/s

Therefore, the initial speed of the bullet is approximately 39.15 m/s.