Answer :
A. Using the equations
"V final(Vf) = V initial(Vo) + acceleration(a) * time(t)
delta x(or y) = Vot + 1/2at^2
Vf^2 = Vo^2 + 2adelta(x)"
Vf is 0
Vo is 4
a is -9.8
delta y = is missing
Vf^2 = Vo^2 + 2adelta(x)
0 = 16 + 2(-9.8)delta y
y= .816
C= 2.316m
Vf^2 = 2 (9.8) 2.316
Vf = 6.738 m/s
B.
V final(Vf) = V initial(Vo) + acceleration(a) * time(t)
Vf = part B
Vo = 0
a = 9.8
t = ?
t = .687s from the top to the bottom but you have to account for it going up too
C.
And using the equation again should you get:
t = .408s
total = 1.096s
"V final(Vf) = V initial(Vo) + acceleration(a) * time(t)
delta x(or y) = Vot + 1/2at^2
Vf^2 = Vo^2 + 2adelta(x)"
Vf is 0
Vo is 4
a is -9.8
delta y = is missing
Vf^2 = Vo^2 + 2adelta(x)
0 = 16 + 2(-9.8)delta y
y= .816
C= 2.316m
Vf^2 = 2 (9.8) 2.316
Vf = 6.738 m/s
B.
V final(Vf) = V initial(Vo) + acceleration(a) * time(t)
Vf = part B
Vo = 0
a = 9.8
t = ?
t = .687s from the top to the bottom but you have to account for it going up too
C.
And using the equation again should you get:
t = .408s
total = 1.096s