A rock is thrown upward with a velocity of 18 meters per second from the top of a 37 meter high cliff, and it misses the cliff on the way back down. When will the rock be 3 meters from ground level? Round your answer to two decimal places.

Answer 1

height = -4.9t^2 + 18t + 37
So we want to find t when height = 3
3 = -4.9t^2 + 18t + 37
4.9t^2 - 18t - 34 = 0
use the quadratic formula to find t, reject the negative value

Answer 2

V^2 = Vo^2+2gh = 0
18^2-19.6h = 0
h = 16.53 m. above cliff.

V = Vo+gTr = 0
18-9.8Tr = 0
Tr = 1.84 s. = Rise time.

0.5*g*Tf^2 = 37-3
0.5*9.8*Tf^2 = 34
Tf = 2.63 s. = Fall time.

T = Tr+Tf = 1.84+2.63 = 4.47 s. = time to reach 3 m above gnd.

Answer 3

Correction: 0.5*g*Tf^2 = 16.53+(37-3)
0.5*9.8*Tf^2 = 50.53
Tf = 3.21 s. = Fall time.

T = Tr+Tf = 1.84+3.21 = 5.05 s = time to reach 3m above gnd.