A cannonball is fired with an initial velocity of 145 meters per second. The object's distance, d, avobe the ground at any time

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meters-per-second

#1

A cannonball is fired with an initial velocity of 145 meters per second. The object’s distance, d, avobe the ground at any time, t, can be represented by the equation below. When will the cannonball be 600 meters above the ground? d = 145t - 5t2

  • a. The cannonball will never reach 600 meters.
  • b. t = 1 sec, t = 5 sec
  • c. t = 29 sec, t = 120 sec
  • d. t = 5 sec., t = 24 sec.

Answer:

600=145t - 5t2 =>5t2 -145t +600=0 /5=> t^2-29t+120=0 =>t^2–5t-24t+120=0 =>
t(t-5)-24(t-5)=0=>(t-5)(t-24)=0=>t=5sec and t=24sec so answer D