A 32 caliber bullet fired from a revolver with a 3 in long...

A .32-caliber bullet fired from a revolver with a 3-in.-long barrel will have a relatively low muzzle speed of about 200 m/s. If it's shot straight up, neglecting air resistance: (a) what is the peak height the bullet will reach? (b) How fast will it be moving when it returns to the height of the gun? (c) How long will the whole trip take?

Transcript

00:01 We are given that the initial velocity of bullet of bullet let that be you that is equal to 200 meter per second and we are given that the bullet is short it shot straight up straight up so let's suppose this is the bullet and it is shot straight up and it is shot straight up and it is shot straight up and it is shot straight up in this direction with initial velocity 200 meter per second.

00:37 Now let the highest distance traveled by this bullet is h.

00:45 So at the highest point the final velocity becomes zero.

00:50 So by using the third equation of the newton's motion equation, we can write v square is equal to u square minus 2g .s.

01:01 This minus sign we have taken because the gravitational acceleration and the velocity are in opposite direction when the bullet is going up so v is equal to zero so from here i can write h is going to be u square by 2g that means 200 multiply 200 divide 2 multiply 2 multiply value of g is 10 so if you saw this we get 2 000 meter so the bullet will reach a height maximum height of 2 000 meter after that it starts it start going down so this is the highest point and it is coming down after that so here the initial velocity for this motion the initial velocity at this point at highest point will become will become zero and we are going to find the final velocity when the bullet reaches to the bullet so we are going to find the final velocity we so again by using third equation we can write v square is equal to u square this time we are going to take plus 2g h plus 2g so u is equal to 0 so we can be written as under root 2g h that means under root 2 multiply 10 multiply value of h is 2 000 so if you saw this we get 200 meter per second because in the question we are given that there is negligible air resistance that means the bullet will strike to the initial point with the same velocity it was drawn up that means it was drawn with an velocity of 200 meter per second so it will strike back with the same velocity that is 200 meter per second now we are going to find the time taken in the complete trip so we will find the time taken to reach the highest point time taken to reach the highest point the highest point so we will use the first equation that is v is equal to u minus g t so when the bullet is going up then the final velocity is equal to zero at highest point and initial velocity we are given 200 meter per second so put the values we get t is equal to u by g so put the value of u that is 200 divide g is 10 so we are going to get 20 second...

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