# Projectile motion

1. Feb 3, 2009

### KFC

1. The problem statement, all variables and given/known data
Someone ask me a problem found in college physics textbook. It states: An arrow is shot at an angle of 45 degree above the horizontal. The arrow hits a tree a horizontal distance away D=220m, at the same height above the ground as it was shot. Use g for the magnitude of the acceleration due to gravity. Find the time with the arrow is travelling in the air.

2. The attempt at a solution

Assum the total time when the arrow travel in the air is T. First, the let the initial velocity be V and the initial horizontal velocity is $$V_{x0}$$ and vertical velocity is $$V_{y0}$$. We have

$$V_{x0}=V_{y0} = D/T$$

For y position (height), when the arrow hit the tree, we have
$$D = V_{y0}T - gT^2/2$$

But $$V_{y0}=V_{x0}$$, this gives $$gT^2/2=0$$ ???

I don't know what's going on here. So I assume the vertical velocity when the arrow hit the tree is ZERO, so

$$0 = V_{y0} - gT = V_{x0} - gT = D/T - gT$$

which gives $$T=\sqrt{D/g}$$

I know the correct answer should be $$T=\sqrt{2D/g}$$,I just don't know what's going on here. I check many times, please tell me where I get the problem wrong.

2. Feb 4, 2009

### LowlyPion

I think that you haven't accounted for the fact that the time to max height is only 1/2*T

Vx = Vy

We know from V = g*t that time to max height is Vy/g = T/2

T = 2*Vy/g

But D = Vx*T

Vx = D/T = Vy

Substituting then

T = 2*Vy/g = 2*Vx/g = 2*D/(g*T)

yields

T2 = 2*D/g

3. Feb 4, 2009

### KFC

Thanks LowlyPion. But I still have two questions.

1) The problem only tells the horizontal distance and vertical distance is the same, from that, how do you know the arrow travel to the highest point when it hit the tree?

2) If I use the equation for distance along y direction, i.e.

$$D = V_{y0}T - \dfrac{1}{2}gT^2$$

I will have $$V_{y0}T = V_{x0}T = D$$, which will lead to $$\dfrac{1}{2}gT^2=0$$, how does this contradiction come form?

3) Someone derive that for me and he said when the arrow hit the tree, the velocity along the y direction is $$V_y = -V_{y0}$$, so

$$D = V_{y0}T - \dfrac{1}{2}gT^2$$

In that case, we will have the correct answer. But why the final velocity along y direction is $$-V_{y0}$$?

4. Feb 4, 2009

### LowlyPion

I think you've misread the problem.

The initial angle is 45°. This means that the initial components of velocity are the same. (And final as it turns out - same height)
Not that the arrow rises to the same height that it travels horizontally - because that's not the case.

D ≠ hmax

As to your Vy, the fact that it is a uniformly accelerated gravity field tells you that whatever vertical speed it had when shot from that height, it will have when it is at that heights again when it falls.