# Determine the time the arrow is in flight

• candycooke
In summary, in order to find the two possible angles at which an arrow can be aimed to travel 30 meters with an initial speed of 25 m/s, you need to set up equations of motion for both the x and y directions. This will result in two equations with two unknown variables, the angle and time. Solving for both variables will give you the two possible angles, with one of them being 33.6 degrees. However, the equation for the y direction of motion provided in the conversation is incorrect and should be corrected to account for the arrow reaching its highest point. Using this correction, the equation for time should be t=2(0-25sin\theta)/-9.8m/s^2,
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.

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?

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.

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.

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.

## 1. How is the time of flight for an arrow determined?

The time of flight for an arrow is determined by measuring the time it takes for the arrow to travel from the point of release to the point of impact. This can be done using a stopwatch or other time-measuring device.

## 2. What factors affect the time of flight for an arrow?

The time of flight for an arrow can be affected by several factors, including the initial velocity of the arrow, the angle at which it is released, air resistance, and the weight and shape of the arrow.

## 3. Can the time of flight for an arrow be calculated?

Yes, the time of flight for an arrow can be calculated using the equation t = √(2d/g), where t is the time of flight, d is the vertical distance traveled by the arrow, and g is the acceleration due to gravity.

## 4. Is there a maximum time of flight for an arrow?

Yes, there is a maximum time of flight for an arrow, which is determined by the initial velocity and angle of release. The maximum time can be calculated using the equation t = 2u*sin(θ)/g, where u is the initial velocity and θ is the angle of release.

## 5. How can the time of flight for an arrow be used in archery?

The time of flight for an arrow is an important factor in archery as it can help determine the accuracy and distance of a shot. By understanding the time of flight, archers can make adjustments to their technique to improve their aim and ultimately hit their target more effectively.

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