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?

h(t) = at2 + vt + h0
0.05 s
2.03 s
3.80 s
3.70 s

Answer 1

C or 3.80 on edge

Explanation:

because

Answer 2

Formula is:h ( t ) = a t² + v t + h o,where: a = - 16 ft/s², v = 60 ft/s and h o = 3 ft.When the toy rocket reach the ground: h ( t ) = 0.- 16 t² + 60 t + 3 = 0t 1/2 = ( - 60 +/- √( 60² - 4 · 3 · ( -16 ) ) / ( -32 )t 1/2 = ( - 60 +/- √(3600 + 192 ) ) / ( - 32 )t = ( - 60 - 61.58 ) / ( - 32 ) - another solution is negative.t = ( - 121.58 ) : ( - 32 ) = 3.80 sAnswer: C ) 3.80 s