A bullet shot directly into a lake is stopped by the water. It takes 1.70x10^-2

Question

A bullet shot directly into a lake is stopped by the water. It takes 1.70x10^-2 seconds for the water to stop the bullet (assume constant acceleration).
A) If the bullets muzzle velocity is Vo=295 m/s, at what water depth would you be safe from this bullet?
B) If, instead the bullet has some other initial velocity Vo that would have taken 2.60x10-2 seconds to stop at that same acceleration, then what would the bullet's speed be when it hit you at the depth determined in part (a)? A bullet shot directly into a lake is stopped by the water. It takes 1.70x10^-2 seconds for the water to stop the bullet (assume constant acceleration).
A) If the bullets muzzle velocity is Vo=295 m/s, at what water depth would you be safe from this bullet?
B) If, instead the bullet has some other initial velocity Vo that would have taken 2.60x10-2 seconds to stop at that same acceleration, then what would the bullet's speed be when it hit you at the depth determined in part (a)? A bullet shot directly into a lake is stopped by the water. It takes 1.70x10^-2 seconds for the water to stop the bullet (assume constant acceleration).
A) If the bullets muzzle velocity is Vo=295 m/s, at what water depth would you be safe from this bullet?
B) If, instead the bullet has some other initial velocity Vo that would have taken 2.60x10-2 seconds to stop at that same acceleration, then what would the bullet's speed be when it hit you at the depth determined in part (a)?
A) If the bullets muzzle velocity is Vo=295 m/s, at what water depth would you be safe from this bullet?
B) If, instead the bullet has some other initial velocity Vo that would have taken 2.60x10-2 seconds to stop at that same acceleration, then what would the bullet's speed be when it hit you at the depth determined in part (a)?

Explanation / Answer

Here ,

time taken to stop bullet ,t = 0.017 s

a)

initial speed , vo = 295 m/s

let the acceleration of the bullet is a

Using first equation of motion

v = u + a * t

0 = 295 + 0.017 * a

a = -17353 m/s^2

Using third equation of motion

d = (295)^2/(2 * 17353 )

d = 2.51 m

the water depth is 2.51 m

b)

let the final speed is v

initial speed , v0 = a * t

v0 = 17353 * 0.026 m/s

v^2 - (17353 * 0.026)^2 = -2 * 17353 * 2.51

v = 341 m/s

the speed at this depth is 341 m/s

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