-0.9 45t An object is thrown verically pard from the surace of a celestial body
ID: 2888185 • Letter: #
Question
-0.9 45t An object is thrown verically pard from the surace of a celestial body at a velocity of 45 meters per second. ts distance from the surface att seconds is given by s a. How fast is the object moving 3 seconds after being thrown? b. How long after the object is thrown does it reach its maximum height? c. How high will the object go? a. The object's velocity after 3 seconds is m/sec. (Round to the nearest tenth as needed.) b. It takes the object sec to reach its maximum height. (Round to the nearest tenth as needed.) c. The object reaches a maximum height ofm. (Round to the nearest tenth as needed.)Explanation / Answer
We have given s(t)=-0.9t^2+45t
a) the velocity of an object is v(t)=d(s(t))/dt=-1.8t+45
plug t=3 into v(t)
v(3)=-1.8*3+45=-5.4+45=39.3
The object's velocity after 3 seconds is 39.3 m/sec
b) it's maximum height v(t)=0
-1.8t+45=0
1.8t=45
t=45/1.8=25
it takes the object 25 seconds to reach it's maximum height
c) We have s(t)=-0.9t^2+45t
substituting t=25 into s(t)
s(25)=-0.9(25)^2+45*25
=-562.5+1125
=562.5
The object reaches a maximum height of 562.5 m
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