Atoy rocket is launched straight up into the air with an initial velocity of 60 ft/s from a table 3 ft above the ground. if acceleration due to gravity is –16 ft/s2, approximately how many seconds after the launch will the toy rocket reach the ground?

7,55 seg

Step-by-step explanation:

Initial Velocity = 60 ft/s

High = 3ft

Acceleration = -16 ft/s2

According to the next formula

H = Vi(t) - 1/2 gt2

We got a cuadratic formula which roots are

t1 = -0,04 seg and t2 = 7,55 seg

the time not negative so t= 7,55 seg

answer: a clock

step-by-step explanation:

step-by-step explanation:

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