1. What is the acceleration in m/s^2 of a car that slows from 50mi/hr to 25 mi/hr in 5 minutes?
1 answer:
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Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.