2 = 1/2 g t^2 + 18 t + 27
0 = -4.9 t^2 + 18 t + 25
use the quadratic formula to solve
... the negative solution is not realistic
A rock is thrown upward with a velocity of 18 meters per second from the top of a 27 meter high cliff , and it misses the cliff on the way back down. When will the rock be 2 meters from ground level?
2 answers
a = -g =-9.81 m/s^2
v = Vi + a t = 18 - 9.81 t
h = Hi + Vi T + (1/2) a t^2 = 27 + 18 t - 4.9 t^2
so
2 = 27 + 18 t - 4.9 t^2
or
4.9 t^2 - 18 t - 25 = 0
solve quadratic, use positive time
I get t = 4.75 seconds or -1.07 which was before the throw
v = Vi + a t = 18 - 9.81 t
h = Hi + Vi T + (1/2) a t^2 = 27 + 18 t - 4.9 t^2
so
2 = 27 + 18 t - 4.9 t^2
or
4.9 t^2 - 18 t - 25 = 0
solve quadratic, use positive time
I get t = 4.75 seconds or -1.07 which was before the throw
2 answers