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

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 s C ) 3.80 s

Look at this and see if it

It equals 1373.5 that’s the solution