# A 0.250 kg skeet (clay target)

A 0.250 kg skeet (clay target) is fired at an angle of ? = 32.2°to the horizon with a speed of vo = 32.3 m/s, as seen in figurebelow. When it reaches the maximum height, it is hit from below bya 15 g pellet traveling vertically upward at a speed of v = 194m/s. The pellet is embedded in the skeet. How much higher did theskeet go up? How much extra distance, ?x, does the skeet travelbecause of the collision?

First Part –

Consider the ideal situation, means there is no air drag.

Now, at the highest point, the y velocity = 0m/s.Using momentum,(0.25)(0) + (15/1000)(194) = (0.015+0.25)(v)

=> v = (0.015 x 194) / 0.265 = 10.98 m/suse the expression –

v^2 = u^2 + 2as0 = 10.98^2 + 2(-9.8)(s)

=> s = 10.98^2 / (2 x 9.8) = 6.15 m higher up.

Second Part –

X- component of velocity = 32.3 cos 32.2 = 27.3 m/s

Y – component of velocity = 32.3 sin32.2 = 17.2 m/s

Now, time taken by the skeet to cover extra height s = 6.15m

s = ut + (1/2)*a*t^2

=> 6.15 = 10.98t – 0.5*9.8*t^2

=> 4.9t^2 – 10.98t + 6.15 = 0

=> t = [10.98 + sqrt(10.98^2 – 2*4.9*6.15)] / (2*4.9) =[10.98 + 0.14] / 9.8 =1.13 s

other value of t = [10.98-0.14] / 9.8 = 1.11 s

So, the extra distance travelleed by the skeet, delta x = 1.13 x27.3 = 30.85 m

and the other value of delta x = 1.11 x 27.3 = 30.3 m

so, our answers are 30.85m or 30.3 m.

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