Answer:
322.6 m/s
Explanation:
Given that there are two components of position;
x= vot
y component is;
y= 1.9 - gt^2/2
When the bullet hits the ground, y=0
1.9 -gt^2/2 =0
Where g = 10ms^-2
t= √2 × 1.9/10
t= 0.62 secs
Therefore,
x= vot
vo = x/t
Where, x= 200 m
vo= 200/0.62 =
vo= 322.6 m/s
Hi there!
We can begin by calculating the distance remaining after the reaction time.
Δd = vt
Calculate the distance traveled within this time:
Δd = (16)(.79) = 12.64 m
Subtract from the total distance:
150 - 12.64 = 137.66 m remaining
We can use the following equation to solve for the acceleration necessary:
vf² = vi² + 2ad, where vf = 0 since the train will have slowed down to rest.
Rearrange in terms of "a":
0 = vi² + 2ad
(-vi²) = 2ad
(-vi²)/2d = a
Plug in the given values:
(-(16²))/2(137.66) = a
-256/275.32 = -.9298 m/s²
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