Calculating Maximum Height and Horizontal Range of 0.25kg Skeet

[wing_88]

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
..

 

[tiny-tim]

Welcome to PF!

Hi wing_88! Welcome to PF!

at highest v=0 …

Nooo.

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

 

[tomcue]

1) To calculate the maximum height, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (25 m/s), and a is the acceleration due to gravity (-9.8 m/s^2). We can rearrange the equation to solve for s, the displacement or maximum height. Plugging in the values, we get s = (0^2 - 25^2)/(-2*9.8) = 3.19 m. Therefore, the skeet went up an extra 3.19 meters before being hit by the pellet.

2) To calculate the extra horizontal distance, we can use the equation x = ut + 1/2at^2, where x is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Since the horizontal velocity remains constant, we can use the formula x = vt, where v is the horizontal velocity (25 m/s) and t is the time. To find the time, we can use the equation v = u + at, where u is the initial velocity (25 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and v is the final velocity (0 m/s). Solving for t, we get t = 25/9.8 = 2.55 seconds. Plugging this value into the formula x = vt, we get x = 25*2.55 = 63.75 m. Therefore, the skeet traveled an extra 63.75 meters horizontally before being hit by the pellet.