Ans:
1)
a) di = 3 m
df = 0 m
g = -9.8 m/s2
vi = 0
So,
df = di + vi(t) + ½(gt2)
t = 2.21 s
b) vf = d/t = 7/2.21 = 9.48 m/s
2)
a) when projectile is in the air, it rises and then falls to a final possition 3 m lower than its starting altitude. We can find the time by using
y = y0 + v0yt – 1/2gt2
where y0 = 0
y = -3 m
Initial vertical velocity
V0y = v0sinϴ = 6.68 m/s
Substituting known values yields
-3 = 6.68 t – 4.9 t2
Solving this eq.
We get
T = 1.7193 s
b) Horizontal displacement = v*t = 11.4886 m
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