Asked by adeeba

a ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity 80 m/ sec .the distance s of the ball from the ground after t seconds is s= 96+80t-4.9 t^2.after how many seconds does the ball strike the ground

Answers

Answered by Reiny
you want height to be zero, so
0 = 96+80t -4.9t^2
using the formula ...
t = (-80 ± √8281.6)/-9.8
= appr 17.45 seconds or a negative
Answered by Anonymous
4r3435
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