A shell is shot with an initial velocity v0 of 42 m/s, at an...

A shell is shot with an initial velocity v_0 of 42 m/s, at an angle of ?_0 = 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

Transcript

00:01 Hi, in this question, the initial velocity v0 is given as 42 meters per second.

00:10 The angle of projection, theta not, is given as 60 degrees.

00:17 We need to determine the horizontal distance of the second fragment.

00:22 The horizontal distance, let us say x, we need to determine.

00:27 Let us find out the x and o coordinates of the highest point using the equation.

00:33 Of motion.

00:34 Let us use the equation of motion and determine the x coordinate.

00:40 We can say x0 is equal to b0 into sine theta not, sine theta not, sine theta not, sine theta not, divided by g.

01:01 Now let us plug the given values and determine the x coordinate.

01:05 V0 is given us 42 meters per second 42 meters per second sine 60 multiply with cost 60 divided by g that is the acceleration due to gravity on calculation we obtained the value of x is equal to x0 is equal to 77 .94 meters now let us determine the y -not value that is the vertical height value y -not is equal to b0 square by 2g multiply with sine square theta x equal to now let us plug the given values v0 square is 4d2 square divided by 2 in 9 .8 multiply with side square 60.

02:21 On calculation we obtained the value of y0 is equal to 67 .5 meters.

02:31 Let us apply the principle of conservation of momentum and find the velocity of the fragment.

02:38 M b .0 this velocity is the original velocity.

02:45 42 meters per second.

02:46 Multiply with cost theta not is equal to m by 2 times v0.

02:58 This v0 is the velocity of the fragment and this v0 is the initial velocity...