An arrow is shot vertically upward. Three seconds later, it is at a height of 35m.?

Question

A) What was the initial speed of the arrow?
B) How long is the arrow in flight from launch to returning to the initial height?( assuming the arrow falls vertically downward.)

Answer 1

Assuming the arrow is just a projectile with no aerodynamics involved.

A) rise = v0 t - 1/2gt^2

v0 = rise/t + 1/2 gt.

B) 0 = v0t - 1/2gt^2

total time = 2v0 / g

= 2 rise / gt + t

Answer 2

Even though the arrow has an upward velocity, it has a downward acceleration,
Up is negative, down is positive!
So plugging in the formula
d = (vi x t) + (0.5 x a x t x t)

where
d is the distance (-35 m)
vi is the initial velocity (what we're looking for)
t is the time (3 sec), and
a is the grav acceleration (9.8 m/s/s)

-35 = vix3 + 0.5x9.8x3x3

-35 = 3vi + 44

vi = (-35-44)/3 = - 50 m/s (neg, b/c it's up)

part b)
we will calculate how long it takes the arrow to reach it's apex (where v = 0) and then double the time.

vf = vi + (a x t)
where
vf is the final velocity, which we will set to zero (i.e when the arrow reaches it's apex), then we can

0 = -50m/s + (9.8 x t)
50 m/s = 9.8 t
50/9.8 = t = 5 secs to reach the apex.
Double that for the round trip - 10 secs.

Answer 3

A) Use the equations for a body undergoing constant acceleration; in this case the acceleration due to gravity, which is what is slowing the arrow after release.

s = ut - g(t^2)

where s is the distance travelled, g is the acceleration due to gravity = 9.8m s-2, and u is the initial speed.
Put in the values given in the question, and solve the equation for u.

B) We use the same equation, but now the difference is we want the time when it's back to it's starting position, or in other words, when s = 0. This represents when the arrow has fallen back to its starting point. This time, we solve for t.

Answer 4

A)s=ut+1/2 at^2
35=3u-5x9
3u=80
u=80/3m/s.
B)0=80/3 t-5t^2
0=16/3 t-t^2
t=0Or 16/3 s answer

Answer 5

y = V0t - gt^2/2

V0 = [y - gt^2/2]/t plug in values for y = 35, t = 3 sec, and you should know g in m/sec / sec - this gives you V0, initial velocity

part 2

set y = 0 = [V0 - gt/2]t which happens when t = 0 and when V0= gt/2 - plug in the values

Answer 6

35/3 = 11.67m/s