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geniusboy [140]
3 years ago
7

a man stands on a flat surface and shoots an arrow vertically into the sky at avelocity of 60 meters per second. calculate the m

aximum height the arrow reached. what was the velocity of the car when it hit the ground?

Physics
1 answer:
Contact [7]3 years ago
6 0
A) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
d=?

Vf^2=Vi^2+2ad
(0m/s)^2=(60m/s)^2+2(-9.81m/s^2)d
0=3600+(-19.62)d
-3600=(-19.62)d
(-3600)/(-19.62)=(-19.62)d/(-19.62)
d=183.5m

c) Vi= 60 m/s
a= -9.81 m/s^2
Vf= 0m/s^2
t=10s
d=?

d=Vi*t+(1/2)a*t^2
d=60m/s(10s)+(1/2)(-9.81m/s^2)(10s)^2
d=600+(-4.905)(100)
d=600+(-490.5)
d=108.5m
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By definition we know that the change in entropy is given by

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