An object is thrown vertically from a height of 6 m6m at 1 m/s1ms. How long will it take for the object to hit the ground?

1 Answer
Feb 10, 2016

t=1,0087 st=1,0087s
t=0,872 s#

Explanation:

"if object moves downward:"if object moves downward:
h=v_i*t+1/2*g*t^2h=vit+12gt2 ;;6=1*t+1/2*9,81*t^26=1t+129,81t2
6=t+4,905*t^26=t+4,905t2
4,905*t^2+t-6=04,905t2+t6=0;ax^2+bx+c=0ax2+bx+c=0
Delta=sqrt(b^2-4*a*c) ;Delta=sqrt(1+4*4,905*6)
Delta =sqrt(118,72)
Delta=10,896
color(red)(t=(-b-Delta)/2*a)" time is not negative "
color(green)(t=(-b+Delta)/2*a)
t=(-1+10,896)/(2*4,905);;t=(9,896)/(9,81)
t=1,0087 s
"if object moves upward:"
t_u=v_i/g=t_u=1/(9,81)
t=0,1019 s" elapsed time while object moves upward"
h_u=V_i^2/(2*g)=1/(19,62)=0,051 m " height from starting point"
h=6+0,051=6,051 m " height to ground"
h=1/2 *g*t^2 " freely falling"
6,051=1/2*9,81*t^2 ; t=sqrt((2*6,051)/(9,81)) t=0,872 s