A rocket is shot straight up into the air from a launch pad on the ground at a rate of 55 feet per second the rocket is rocket is tracked by a rangefinder (on the ground) that is 478 feet from the launch pad let d represent the distance from the rocket to the rangefinder and t represent the time in seconds since blastoff answers

d= 501.36 feet

 t = 5.5 seconds

Step-by-step explanation:

Where U = Initial velocity, 55ft/s

            d = ?

            t=  ?

From the solution diagram at Hmax the  Final velocity of the rocket is zero. The equation of motion gives :

Hmax = U²/2g  

          = 55²/2 (10)

          = 151.25ft

Using Pythagora's  theorem:

d²= H²max +478²

   =  151.25² + 478²

   = 251,360.5625

d =√251,360.5625

  = 501.3587

  = 501.36 ft

The time t to reach Hmax is given by:

t = U/g

  = 55/10

  =5.5 seconds

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.