# Determine the time the arrow is in flight

1. Feb 17, 2008

### candycooke

An arrow is shot with an initial speed of 25 m/s (and lands at the same height as its launch height). There are two possible angles at which it chould be aimed so that it goes 30 meters. What are they?

Givens:
v(i)=25m/s
a=-9.8m/s^2
d(x)=30m

I know the first step is to determine the time the arrow is in flight, but I'm not exactly sure which formula to use in order to find time.

2. Feb 17, 2008

### G01

In this problem you have two unknown variables, the angle,$$\theta$$, and the time, t. You need to find two equations involving both of those variables, then you can solve for both of them.

HINT:

You have motion in two directions here. Can you set up an equation of motion for each direction involving $$\theta$$ and t?

3. Feb 17, 2008

### candycooke

since the porabola is symetrical, than v(i)=v(f)
for y components:
t=v(f)-v(i)a
=0a

for x components:
t=d/v
=30m/25cos$$\theta$$

therefore:0=30m/25cos$$\theta$$
$$\theta$$=33.6

So I've figured out one angle but I am unsure how to determine the second possible angle.

4. Feb 17, 2008

### G01

Your equation for the x direction is correct. Leave it as is.

Your equation for the y direction of motion is wrong. Think about what it says. How can the arrow take 0 seconds to make it's trip?

You solved for t incorrectly. Check your work. Also, I suggest using the highest point as your final point in the y direction. (Why you ask? What is v_fy at the highest point?) Remember that the arrow will hit it's highest point at t/2 seconds, where t is the time of the whole trip.

5. Feb 18, 2008

### candycooke

Since v_fy=0 at the highest point:
t/2=(0-25sin$$\theta$$)/-9.8m/s^2
therefore:
t=2(0-25sin$$\theta$$)/-9.8m/s^2

but this is not equal to t=30m/(25cos$$\theta$$)

I am so confused.