A Gun is held horizontally at a target 60 m away, and the sight is exactly align

Question

A Gun is held horizontally at a target 60 m away, and the sight is exactly aligned with the barrel. The bullet hits the target 2.5 cm below the point where the barrel was aimed. Assume for the moment that there is no air resistance (drag), even though that is not a realistic assumption for a bullet. (We’ll consider the effect of drag in part d.)

(a) What was the bullet's flight time?
(b) What was the bullet's speed as it left the barrel (i.e. the "muzzle velocity")?
(c) How far away would the target have to be for the bullet to drop 25 cm vertically along the way?

(d) Now consider the original 2.5 cm drop for the target at 60 m, and add some realism back by assuming that air resistance IS present (but there is no wind) when interpreting the 2.5 cm measurement. Does this imply that the bullet's true muzzle velocity was higher, lower, or the same as your estimate in part (b)? Explain your reasoning.

Explanation / Answer

a) time taken to fall 2.5cm due to gravity can be calculated by

h = (1/2)gt2

t= sqrt(2h/g)

plugging in the values we get

t = 0.07 s

bullets flight time t = 0.07 s

b) speed = distance/time

v = 60/0.07 = 857.1 m/s

c) t= sqrt(2h/g)

then time of flight would be

t = 0.226 s

distance of target

X =vt

X = 193.6m

d) the speed should be greater so that it takes same time to reach the target even in presence of drag

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