Nick throws a ball downward off a building Vi = -26.82 m/s
a = -9.8 m/s/s
t = 1.7 s
What if the balls’ displacement was – 42 m. Keep initial velocity the same. Find time and final velocity??

6.74 s

-39.27 m/s

Explanation:

Given:

Δy = -42 m

v₀ = -26.82 m/s

a = -9.8 m/s²

Find: t and v

Δy = v₀ t + ½ at²

-42 m = (-26.82 m/s) t + ½ (-9.8 m/s²) t²

-42 = -26.82t − 4.9t²

4.9t² + 26.82t − 42 = 0

t = [ -26.82 ± √(26.82² − 4(4.9)(-42)) ] / 2(4.9)

t = (26.82 ± 39.27) / 9.8

t = 6.74 s

v² = v₀² + 2aΔy

v² = (-26.82 m/s)² + 2 (-9.8 m/s²) (-42 m)

v = -39.27 m/s

the answer to this question is 60 n

hèllo, sorry i really need some more points so if you need this qustion answered than retype it