A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maximum height, it is hit from below by a 15 g pellet travelling upward at a speed of 200 m/s. The pellet is embedded in the skeet. (a) How much higher did the skeet go up? (b) How much extra distance does the skeet travel because of the collision?

the image is here:
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thanks :)

so a.) 6.5m

in the b.) is the answer 10.07m?

coz i'm not really sure with what i did. but i solved for t using x=v1t+1/2at^2.. then multiplied it with 30cos30. :O

Oh, I see! So we have a skeet flying through the air, minding its own business, when suddenly it gets hit from below by a speedy pellet. Let's calculate the fun details!

(a) To find out how much higher the skeet went up, we need to determine the maximum height it reached before the collision. We can use the projectile motion equations for this. Since we're given the initial velocity (25 m/s) and the launch angle (30 degrees), we can use the equations to find the maximum height.

Using the equation for maximum height, hmax = (v^2 * sin^2θ) / (2g), where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity, we plug in the values:

hmax = (25^2 * sin^2(30)) / (2 * 9.8)

Calculating that, we find hmax ≈ 5.06 meters.

So, the skeet went up approximately 5.06 meters before it was hit by the pellet.

(b) Now let's determine the extra distance the skeet travelled due to the collision. Since the pellet is embedded in the skeet, we can treat their collision as an inelastic collision. In an inelastic collision, the objects stick together and move as one.

To calculate the extra distance, we need to first find the final velocity of the combined skeet and pellet after the collision. We can use the conservation of momentum to do that. The initial momentum is the sum of the individual momenta, and the final momentum is the momentum of the combined object. We can use the equation:

m1 * v1 + m2 * v2 = (m1 + m2) * vf

For this, let's consider m1 as the mass of the skeet and m2 as the mass of the pellet.

m1 = 0.25 kg (mass of the skeet),
v1 = 0 m/s (since we want the extra distance after the collision),
m2 = 0.015 kg (mass of the pellet),
v2 = -200 m/s (negative because it's moving upward).

Solving the equation, we can calculate vf.

0.25 kg * 0 m/s + 0.015 kg * -200 m/s = (0.25 kg + 0.015 kg) * vf
-3 kg·m/s = 0.265 kg * vf
vf ≈ -11.32 m/s

So, the final velocity after the collision is approximately -11.32 m/s. Since the velocity is negative, it means the object is moving downward.

Now, to find the extra distance, we can use the equation for distance traveled when an object is accelerating:

d = (vf^2 - vi^2) / (2a)

Since we want to calculate distance, we can assume initial velocity vi as 0 m/s and acceleration a as the acceleration due to gravity (-9.8 m/s^2).

Plugging in the values:

d = (-11.32^2 - 0) / (2 * -9.8)
d ≈ 6.62 meters

So, the skeet traveled an additional distance of approximately 6.62 meters due to the collision.

Voila! We've calculated how much higher the skeet went up and how much extra distance it traveled because of the collision. Physics can be fun too, right?

To solve this problem, we can use the principles of projectile motion and conservation of momentum.

(a) To calculate how much higher the skeet goes up, we need to find the maximum height it reaches before the collision. In projectile motion, the projectile (skeet) follows a parabolic trajectory.

First, let's break the initial velocity of the skeet into its horizontal and vertical components.

Horizontal Component:
v_0x = v_0 * cos(theta)
v_0x = 25 m/s * cos(30 degrees)
v_0x = 21.65 m/s

Vertical Component:
v_0y = v_0 * sin(theta)
v_0y = 25 m/s * sin(30 degrees)
v_0y = 12.5 m/s

The time taken for the skeet to reach its maximum height can be calculated using the vertical component of its velocity and the acceleration due to gravity.

v_fy = v_0y + a_y*t

At maximum height, the vertical component of velocity becomes zero. So, v_fy = 0. Therefore:

0 = 12.5 m/s + (-9.8 m/s^2) * t

Solving this equation, we can find the time taken (t) for the skeet to reach its maximum height.

Next, we can calculate the maximum height (h) using the equation of motion:

h = v_0y * t + 0.5 * a_y * t^2

Plug in the values of v_0y, t, and a_y to obtain the value of h.

(b) To find the extra distance the skeet travels due to the collision, we need to consider the conservation of momentum.

The momentum of the system (skeet + pellet) before the collision is equal to the momentum of the system after the collision.

Initial Momentum (before collision):
P_initial = m_skeet * v_skeet_initial + m_pellet * v_pellet_initial

Final Momentum (after collision):
P_final = (m_skeet + m_pellet) * v_final

Here, m_skeet is the mass of the skeet, m_pellet is the mass of the pellet, v_skeet_initial is the velocity of the skeet before collision, v_pellet_initial is the velocity of the pellet before hitting the skeet, and v_final is the final velocity of the system after the collision.

From the given information, we know the mass of the skeet (0.25 kg) and the mass of the pellet (15 g = 0.015 kg), as well as the initial velocity of the skeet (25 m/s) and the initial velocity of the pellet (200 m/s).

Solve the conservation of momentum equation to find the final velocity (v_final) of the system.

Once you have the final velocity, you can calculate the distance (d) travelled by the skeet using the equation of motion:

d = v_final * t

where t is the time taken for the skeet to reach its maximum height that we calculated earlier.

Plug in the values to find the extra distance travelled by the skeet due to the collision.

a.extra height:

Momentum before collision=momentium after collision
but the clay target has no upward momentum at the top, so
.015*200=(.015+.25)V solve for V, that is the new upward velocity.

1/2 mv^2=mgh
h=1/2 v^2/g

now for the extra distance. Figure the time it takes to go h
0=V-9.8t solve for t, then double it (goes back down), that is the extra time in air.
extra distance=2t*25cos30