a rocket is fired straight upward from ground level with an initial velocity of 480Km)hr. After t second , its distance above the ground level is given by 480t-16t^2. For what time interval is the rocket more than 3200Km above ground level?​

answer

Explanation:

The rocket is 19.5 m above the ground when h=19h=19:

h19.50tttamp;=−8t2+32tamp;=−8t2+32tamp;=−8t2+32t−19.5amp;=−32±322−4(−8)(−19.5)−−−−−−−−−−−−−−−−√2(−8)amp;=−32±1024−624−−−−−−−−−√−16amp;=−32±400−−−√−16amp;=−32±20−16amp;=−32+20−16=0.75amp;=−32−20−16=3.25amp;amp;amp;amp;amp;amp;amp;amp;Substitute h=19.amp;Subtract 19.5 on both sides.amp;Solve for t using the Quadratic Formula.amp;Simplify.amp;Subtract.amp;Evaluate the root.amp;Find the two solutions.

h

19.5

0

t

t

t

​  

amp;=−8t  

2

+32t

amp;=−8t  

2

+32t

amp;=−8t  

2

+32t−19.5

amp;=  

2(−8)

−32±  

32  

2

−4(−8)(−19.5)

​  

​  

amp;=  

−16

−32±  

1024−624

​  

​  

amp;=  

−16

−32±  

400

​  

​  

amp;=  

−16

−32±20

​  

amp;=  

−16

−32+20

​  

=0.75

amp;=  

−16

−32−20

​  

=3.25

​  

amp;

amp;

amp;

amp;

amp;

amp;

amp;

​  

amp;Substitute h=19.

amp;Subtract 19.5 on both sides.

amp;Solve for t using the Quadratic Formula.

amp;Simplify.

amp;Subtract.

amp;Evaluate the root.

amp;Find the two solutions.

​  

The rocket will then be at least 19.5 m above the ground for the interval 0.75≤t≤3.25  

0.75≤t≤3.25

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