Certain rifles can fire a bullet with a speed of 965 mathrmm...

Certain rifles can fire a bullet with a speed of 965 $\mathrm{m} / \mathrm{s}$ just as it leaves the muzzle (this speed is called the muzzle velocity). If the muzzle is 70.0 $\mathrm{cm}$ long and if the bullet is accelerated uniformly from rest within it, (a) what is the acceleration (in $g^{\prime}$ s) of the bullet in the muzzle, and (b) for how long (in ms) is it in the muzzle? (c) If, when this rifle is fired vertically, the bullet reaches a maximum height $H,$ what would be the maximum height (in terms of $H$ ) for a new rifle that produced half the muzzle velocity of this one?

Transcript

00:01 In this question, we are given that a certain rifle can fire a bullet with speed of 95 meter per second just as it leaves the muzzle and the muzzle is 70 centimeter long.

00:12 And the bullet is accelerated uniformly from wrist.

00:15 Then what is the acceleration in terms of gs in the bullet? so we are given that this is the muzzle and this is where the bullet is initially and this is 70 centimeters.

00:26 So let's call it 0 .7 meters.

00:28 It leaves the muscle at at least this point at 965 meter per second so we need to find the acceleration so i think we can use the second the third equation of motion which is v square as equal to v0 square plus 2 as and since it accelerates from wrist here we have 965 square here we have zero square plus two times eight times 0 .7 so the value of a comes out as 965 square over one point four so the value of a from here comes out as this is six six five one six zero point seven one and meter per second square but they need in terms of g so we divide this by nine point eight and we are getting the value as sixty seven thousand eight hundred and seventy three point sixty seven thousand eight hundred and seventy three point five four g meter per second square so this is a huge number and then we talk about part b that for how long it is in the muzzle in milliseconds so for how long we need to find the time and the time we can easily find using the first equation of motion that v is equal to v0 plus a t so v is 965 v0 is 0 a 6651 601 and t is as a test so from here, we are getting the value as, which is grab my calculator over here is 965 over on.

02:07 So, which is coming as 1 .45 milliseconds.

02:13 So it's a required time.

02:15 And then we talk about part c that if the muscle is fired vertically, then what is the maximum height to which it will attain? so the bullet is fired from here which is 70 or let's call it 0 .7 meters from here...

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