A 0.25-kg skeet (clay target) is fired at an angle of 30 degrees to the horizon

Question

A 0.25-kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25/ms:

When it reaches the maximum height, it is hit from below by 15-g pellet traveling vertically upward at a speed of 200 m/s. The pellet is embedded in the skeet.

b)How much extra distance, , does the skeet travel because of the collision?

the first part of the question asked how much higher the skeet went up by becouse of the collison which i have worked out to be 6.54 but i cant figure out this last question so nay help would be greatly appricated.

img85.imageshack.us/img85/905/physicsx.jpg (link to the image)

Explanation / Answer

Ok, you know that when an object reaches its maximum altitude then the vertical velocity is 0 m/s, but it still has a horizontal constant velocity. Let's find the maximum altitude : Decomposing the velocity : Vx = 25*cos(30) = 12.5sqrt(3) m/s Vy = 25*sin(30) = 12.5 m/s Using Vy, we will find the maximum altitude : Knowing that the vertical speed at it is 0 ms : Vf^2 = Vo^2 - 2gH Considering g = 10 m/s^2 0 = 12.5^2 - 20H H = 7.8 meters Let's find the time, the skeet took : Vf = Vo -gt 0 = 12.5 -10t t = 1.25 The horizontal distance in this time : X = 1.25*12.5sqrt(3) = 27 meters. Now, in that moment is gonna bt hit by a pellet, so let's use the conservation of linear momentum, but only on the vertical. In other words : Initial momentum in Y = Final momentum in Y 0.25*0 + 15/1000*200 = 0.265*V'y V'y = 11.3 m/s V', is the velocity, that hte skeet and the pellet will have in Y, so the velocity will be : 11.3 m/s in Y, and to find the velocity in X : 0.25*12.5sqrt(3) + 15/1000*0 = 0.265V'x V'x = 20.4 m/s New velocity : Vy' = 11.3 ; Vx = 20.4 m/s Let's find how much higher did the skeet went up : Initial velocity : 11.3 m/s Final velocity must be : 0 m/s g = 10 m/s^2 0 = 11.3^2 - 2*10*H' H' = 6.4 meters Let's find the time the skeet took : 0 = 11.3 -10t' t' = 1.13 The distance : X' = 1.13*12.5sqrt(3) = 24.5 meters So the skeet went up, 6.4 meters more. Now, the skeet is : 6.4 +7.8 = 14.2 meters over the ground, so let's find the time, the skeet is gonna take to hit the ground again : initial velocity = 0m/s g = 10 m/s^2 H = 14.2 m Using : H = Vo*t - g*t^2 / 2 14.2 = 1/2*10*t^2 t = 1.68 seconds distance = 1.68*12.5sqrt(3) = 36.3 Total distance = X + X' + distance : 27 +24.5 +36.3 = 87.8 meters.

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