Question

Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity). The muzzle is 75.0 cm long and the bullet is accelerated uniformly from rest within it. What is the acceleration (in {g}'s) of the bullet in the muzzle? 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?

Answer 1

Answer:

a) By $v^2 = u^2 + 2as$ => a= 70291.70.

(b)By $v = u + at$ => t= 1.58 ms.

(c)By $v^2 = u^2 - 2gh$ => H = 46045.92 m.

Explanation:

a) By $v^2 = u^2 + 2as$

$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$