Answer:
57.97m/s
12.1s
Explanation:
Use one of the key equations of accelerated motion and sub in the values
vf^2=vi^2+2aΔd
vf^2=(20)^2+(2*9.8)(150)
vf^2 = 3340
vf = 57.79m/s (square rooted both sides)
now use another equation
Δd = (vi+vf/2)Δt
and rearrange for Δt
Δt = (2Δd-vf)/vi
sub in values
Δt = (2*150-57.79)/20
Δt = 12.1s
mark me as brainliest if this helped!
