Calculating Maximum Height and Horizontal Range of 0.25kg Skeet

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SUMMARY

The discussion focuses on calculating the maximum height and additional horizontal range of a 0.25 kg skeet shot at a 30-degree angle with an initial speed of 25 m/s, which is subsequently hit by a 15 g pellet traveling vertically at 200 m/s. The key equations involve conservation of momentum during the collision and projectile motion principles to determine the skeet's new trajectory. The conservation of vertical momentum is crucial for calculating the additional height gained after the collision, while the horizontal distance can be derived from the horizontal component of the skeet's velocity and the time of flight.

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wing_88
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Homework Statement


0.25-kg skeet is fred at an angle 30 to horizontal with a speed 25m/s.
when it reaches tha max. height, it is hit from below by a 15g pellet traveling verticaly upwared at speed of 200m/s. the pellet is embedded to the skeet.
1) how much higher did the skeet go up?
2) how much extra horizontal distance?


Homework Equations





The Attempt at a Solution


at highest v=0
vsin 0 is the speed of skeet
horizontal range should be v times t
..
 
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Welcome to PF!

Hi wing_88! Welcome to PF! :wink:
wing_88 said:
at highest v=0 …

Nooo. :redface:

Anyway, start by considering what is conserved at the time of collision … then what is the equation for that? :smile:
 

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