Question

An archer fires an arrow straight up into the air with a speed vo = 42 m/s. Neglect air resistance.

1. Find the maximum height h reached by the arrow, in meters.
2.Write an expression for the time the arrow is in the air until it returns to launch height in terms of known quantities.

Answer 1

You need a direct answer or explanation?

Answer 2

anyway , we use the first law of motion
Vf=Vo-a.t
as it is a deceleration, not acceleration
then t=4.281sec

Answer 3

@3mar would I use that to find the height? or

Answer 4

No sir
the hieght has its own equation which is
Vf^2=Vo^2-a.d

Answer 5

Got it?

Answer 6

@3mar so what would acceleration and distance equal if they only provide initial velocity?

Answer 7

acceleration is the gravitational deceleration -9.81 m/s2 as it go straight up.

Answer 8

The height equation is in the form:
\[d=v.t+0.5a.t^2\]

Answer 9

v is the final velocity
t is the time for complete climb
and a is -9.81 m/s2 as you know
Just get the d.

Answer 10

Is it simple now?

Answer 11

sorry! how would I find time?

Answer 12

from the relation states: Vf=Vo+a.t
Vf=0
Vo=42
a=-9.81

Got it?

Answer 13

This is the climb up time.
Do the same for falling.

Answer 14

Excuse me, I have to go now.
Could you allow?