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3 votes
A bullet is fired straight up from a gun with a

muzzle velocity of 152 m/s.
Neglecting air resistance. what will be its
displacement after 6.9s?

1 Answer

5 votes

Answer: 815.51 m

Step-by-step explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:


V=V_(o)+gt (1)


V^(2)=V_(o)^(2)+2gy (2)

Where:


V is the final velocity of the bullet


V_(o)=152 m/s is the initial velocity of the bullet


g=-9.8 m/s^(2) is the acceleration due gravity, always directed downwards


t=6.9 s is the time


y is the vertical position of the bullet at
t=6.9 s

Let's begin by finding
V from (1):


V=152 m/s-9.8 m/s^(2)(6.9 s) (3)


V=84.38 m/s (4)

Now we have to substitute (4) in (2):


(84.38 m/s)^(2)=(152 m/s)^(2)-2(9.8 m/s^(2))y (5)

Isolating
y:


y=815.511 m This is the displacement of the bullet after 6.9 s

User Lionel Briand
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