Question

A toy rocket is launched straight up into the air with an initial velocity of 60ft/s from a table 3 ft above the ground. If acceleration due to gravity is -16ft/s^2, approximately how many seconds after the launch will the toy rocket reach the ground?

Answer 1

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:
 t = 3.8 s
 option 3