Question
A rok is thrown 24 meters/sec upward and lands in a 10 meter deep hole. How long is the rock in the air until it reaches the bottom of the hole?
Answers
Elena
Upward motion:
v=vₒ-gt
v=0
t= vₒ/g=24/9.8=2.45 s.
The height
h = [(v(fin) ² -v(init)²]/2•(-g) = 24²/2•9.8=29.4 m
Downward motion from the height
H=29.4+10 =39.4 m.
t=sqrt(2H/g) = sqrt(2•39.47/9.8) = 2.84s
Total time = 2.45+2.84 =5.29 s.
v=vₒ-gt
v=0
t= vₒ/g=24/9.8=2.45 s.
The height
h = [(v(fin) ² -v(init)²]/2•(-g) = 24²/2•9.8=29.4 m
Downward motion from the height
H=29.4+10 =39.4 m.
t=sqrt(2H/g) = sqrt(2•39.47/9.8) = 2.84s
Total time = 2.45+2.84 =5.29 s.