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

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
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions