use 2nd equation of motion.
S=UT+1/2ATSQURED
A ball is thrown directly downward with an initial speed of 8.95 m/s, from a height of 29.0 m. After what time interval does it strike the ground?
4 answers
h = Hi + Vi t - 4.9 t^2
0 = 29 - 8.95 t - 4.9 t^2
t = [8.95+/-sqrt(8.95^2 + 4*29*4.9)]/(2*4.9)
t = [8.95 +/-25.46 ] /9.81
use + t
3.51 seconds
0 = 29 - 8.95 t - 4.9 t^2
t = [8.95+/-sqrt(8.95^2 + 4*29*4.9)]/(2*4.9)
t = [8.95 +/-25.46 ] /9.81
use + t
3.51 seconds
the answer is not 3.51 seconds when i type in the answer it doesn't work
it's a quadratic
9.8 t^2 + 8.95 t -29 = 0
Answer works out to 1.32 sec
9.8 t^2 + 8.95 t -29 = 0
Answer works out to 1.32 sec
0 answers