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Question: An Object travels through the air following a path described by h(t) = -4
Question: An Object travels through the air following a path described by h(t) = -4.9t^2 + 17t, where h is the height, in meters, above ground and t is the time in seconds. How long, to the nearest tenth of a second, is the object in the air? Quick help would be greatly appreciated.
Expert Solution
The object is in air for 3.5 seconds.
Step-by-step explanation
The movement of the object is described by the equation
h(t)=−4.9t2+17t
where h is the height in meters and
t is the time in seconds.
We are asked how long is the object in air. Take note that when an object is thrown up, it will remain in air until it falls down and reaches the ground, and when it reaches the ground, h(t)=0.
0=−4.9t2+17t
4.9t2=17t
4.9t=17
t=4.917?
t=3.5 seconds
The object is in air for 3.5 seconds.
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