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: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.