Asked by Bree
An arrow is shot straight upward at 55 m/s.
a) find the time until the arrow reaches its peak
b)how high was the arrow at its peak
c)find the total time the arrow is in the air
d)what was the velocity of the arrow after 3.5 seconds
e)how high was it at 3.5 seconds
a) find the time until the arrow reaches its peak
b)how high was the arrow at its peak
c)find the total time the arrow is in the air
d)what was the velocity of the arrow after 3.5 seconds
e)how high was it at 3.5 seconds
Answers
Answered by
Henry
a. V = Vo + gt.
t = (V-Vo)g.
Tr = (0-55)/-9.8 = 5.6 s. = Rise time.
b. h = Vo*t + 0.5g*t^2.
h = 55*5.61 - 4.9*(5.61)^2 = 154.3 m.
c. Tf = Tr = 5.61 s.
Tr + Tf = 5.61 + 5.61 = 11.22 s. = Time
in air.
d. V=Vo + gt = 55 - 9.8*3.5 = 20.7 m/s.
e. h = Vo*t - 0.5g*t^2
t = (V-Vo)g.
Tr = (0-55)/-9.8 = 5.6 s. = Rise time.
b. h = Vo*t + 0.5g*t^2.
h = 55*5.61 - 4.9*(5.61)^2 = 154.3 m.
c. Tf = Tr = 5.61 s.
Tr + Tf = 5.61 + 5.61 = 11.22 s. = Time
in air.
d. V=Vo + gt = 55 - 9.8*3.5 = 20.7 m/s.
e. h = Vo*t - 0.5g*t^2
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