Question

A toy 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?

Answer 1

Answer:

Answer:

t = 3.8 s

option 3

Step-by-step explanation:

For this case we have the following equation:

 h (t) = at ^ 2 + v * t + h0

 Substituting values we have:

 h (t) = - 16 * t ^ 2 + 60 * t + 3

 We equate the equation to zero:

 -16 * t ^ 2 + 60 * t + 3 = 0

 We look for the roots of the polynomial:

 t1 = -0.04935053979258153

 t2 = 3.7993505397925817

 We are left with the positive root and round:

 t2 = 3.8 s

Answer 2

Answer:

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