Solving for Horizontal Distance of an Arrow Shot Over Level Ground

[Eggphys]

Homework Statement

An archer shoots an arrow over level ground. The arrow leaves the bow at a height of 1.50 m with an initial velocity of 79.5 m/s in a horizontal direction. At what horizontal distance (in m) does this arrow strike the ground? The answer is 44m but I don't know how to get to that answer.

The Attempt at a Solution

I tried to attempt the solution using V=Vo+at to find t then using X=Xo+Vot+(1/2)at^2 to find X but it gives the wrong answer. Not exactly sure where I am going wrong or if I am using the right equations or not.

Any help would be greatly appreciated. Thanks

 

[Fewmet]

Homework Statement

An archer shoots an arrow over level ground. The arrow leaves the bow at a height of 1.50 m with an initial velocity of 79.5 m/s in a horizontal direction. At what horizontal distance (in m) does this arrow strike the ground? The answer is 44m but I don't know how to get to that answer.

The Attempt at a Solution

I tried to attempt the solution using V=Vo+at to find t then using X=Xo+Vot+(1/2)at^2 to find X but it gives the wrong answer.

Welcome to Physics Forums, Eggphys.

The heart of a problem like this is treating perpendicular vectors independently. In this case, that means that the time for the arrow to fall vertically is not affected by its horizontal velocity. The general approach is to find the time to fall and then see how far the arrow goes forward in that time.

Do you know how to do each of those steps? If so, and you are not getting 43.99 m, post your work and we'll help.

 

[Eggphys]

Welcome to Physics Forums, Eggphys.

The heart of a problem like this is treating perpendicular vectors independently. In this case, that means that the time for the arrow to fall vertically is not affected by its horizontal velocity. The general approach is to find the time to fall and then see how far the arrow goes forward in that time.

Do you know how to do each of those steps? If so, and you are not getting 43.99 m, post your work and we'll help.

Thanks for the quick response and cheers for the welcome.
I'm not quite sure how to do each steps. For finding the time to take to fall I used
X=Xo+Vot+(1/2)at^2 then adding in the values
0=1.5+4.9t^2 and getting the answer 0.25
then using the same equation for distance traveled
X=79.5*0.25+4.9*0.06 but i get the answer of 20.17m. Not sure where I'm going wrong.

 

[Fewmet]

Thanks for the quick response and cheers for the welcome.
I'm not quite sure how to do each steps. For finding the time to take to fall I used
X=Xo+Vot+(1/2)at^2 then adding in the values
0=1.5+4.9t^2 and getting the answer 0.25
then using the same equation for distance traveled
X=79.5*0.25+4.9*0.06 but i get the answer of 20.17m. Not sure where I'm going wrong.

(I'm fairly new to the forum and you are the first time I've remembered to offer the welcome I see others do. Your warm response will make it easier to remember next time.)

You have three or four problems in your approach. I have written extensively in hopes that it will model how to think through this and help you avoid repeating the mistakes. (I also am playing it safe by assuming you understand less about doing this that you probably do). I think being patient and careful in reading will be worth the effort.

Consider what X=X_o+V_ot+(1/2)at² is saying. It tells you how to figure out where the arrow will end up (i.e., how far it is from X=0). You take its starting position (X_o), add to it how far it goes because of its initial velocity (V_ot) and then add how far extra it gets because it is accelerating (.5at²).

Now start by identifying where X=0 is (meaning in the vertical direction...we usually call that Y, not X). By writing

0=1.5+4.9t^2

you are implicitly saying that X=0 meters is at the ground (since that is where it ends).

Velocity and acceleration are vectors, so their directions are crucial. That means you must establish the positive and negative directions. It looks like you are choosing down as negative. If so, the arrow starts at X_o= 1.5 m, as you have said. Acceleration, though, is directed downward, so it must be
-9.8 m/s².

That's the first place you went wrong. Also note that you were careless with negative signs and algebra. As you set it up the time would come out to be \sqrt{-.5533} seconds.

For the horizontal motion, you wrote

X=79.5*0.25+4.9*0.06

That is saying that the arrow starts at 0 m, its initial horizontal velocity of 79.5 m/s carries it forward a distance of 79.5*0.25 meters (we've already seen that is the wrong time), and its horizontal acceleration of 9.8m/s² gives it another horizontal distance of 9.8*2.5² meters (again, that 2.5 is incorrect).

Do you see the mistake in substituting in the equation? Fixing that and using the correct time should get you the 44 m.

 

[Eggphys]

(I'm fairly new to the forum and you are the first time I've remembered to offer the welcome I see others do. Your warm response will make it easier to remember next time.)

You have three or four problems in your approach. I have written extensively in hopes that it will model how to think through this and help you avoid repeating the mistakes. (I also am playing it safe by assuming you understand less about doing this that you probably do). I think being patient and careful in reading will be worth the effort.

Consider what X=X_o+V_ot+(1/2)at² is saying. It tells you how to figure out where the arrow will end up (i.e., how far it is from X=0). You take its starting position (X_o), add to it how far it goes because of its initial velocity (V_ot) and then add how far extra it gets because it is accelerating (.5at²).

Now start by identifying where X=0 is (meaning in the vertical direction...we usually call that Y, not X). By writing

you are implicitly saying that X=0 meters is at the ground (since that is where it ends).

Velocity and acceleration are vectors, so their directions are crucial. That means you must establish the positive and negative directions. It looks like you are choosing down as negative. If so, the arrow starts at X_o= 1.5 m, as you have said. Acceleration, though, is directed downward, so it must be
-9.8 m/s².

That's the first place you went wrong. Also note that you were careless with negative signs and algebra. As you set it up the time would come out to be \sqrt{-.5533} seconds.

For the horizontal motion, you wrote

That is saying that the arrow starts at 0 m, its initial horizontal velocity of 79.5 m/s carries it forward a distance of 79.5*0.25 meters (we've already seen that is the wrong time), and its horizontal acceleration of 9.8m/s² gives it another horizontal distance of 9.8*2.5² meters (again, that 2.5 is incorrect).

Do you see the mistake in substituting in the equation? Fixing that and using the correct time should get you the 44 m.

Thanks a lot mate, I see where I went wrong with the algebra. I really appreciate it

 

[Eggphys]

I have another projectile motion problem and instead of making another thread ill just update this one.

1. Homework Statement

At Acapulco, professional divers jump from a 33.1 m high cliff into the sea. At the base of the cliff, a rocky ledge sticks out for a horizontal distance of 6.06 m. With what minimum horizontal velocity (in m/s) must the divers jump off if they are to clear this ledge?

3. The Attempt at a Solution

I tried what I did for the previous problem by adding in the values and solving for T using
0=33.1-4.9t^2 and getting the answer T=2.6
then solving initial Velocity with
6.06=0+2.6Vo-33.12 and getting Vo=10.41

But it feels like I'm missing the final step but I'm unsure what it is? Can anyone shed some light on what's next? Thanks

 

[RobL14]

Okay, so you found the time t that it'd take for the diver to reach y = 0. Now you need to find some horizontal velocity in the x-direction that let's him clear the ledge x = 6.06m in that time, t = 2.6s.

 

[jhae2.718]

6.06=0+2.6Vo-33.12 and getting Vo=10.41

Why do you have 33.12 here?

 

[Eggphys]

Why do you have 33.12 here?

1/2at^2 so 4.9*2.6^2

 

[jhae2.718]

What is the acceleration in the x direction?

 

[Eggphys]

What is the acceleration in the x direction?

Oh, a in the x direction is 0 right? So that make's the equation
6.06=2.6Vo
which gives Initial Velocity of 2.33. Which is correct right?. Haha, I can't believe it was that simply of a mistake. Thanks mate

 

[jhae2.718]

That's what I get, assuming the diver has only a horizontal initial velocity (since the problem doesn't given an angle it's a reasonable assumption).

 

[Eggphys]

That's what I get, assuming the diver has only a horizontal initial velocity (since the problem doesn't given an angle it's a reasonable assumption).

Sweet, I appreciate the help mate.

 

[jhae2.718]

No problemo. Glad you got everything worked out!

 

[subho123]

As the problems discussed here are of projectile motion, that's why i am posting the problem that i have stuck in with...the problem says..

"A stone is thrown with a velocity u from the peak of a Pillar of height h from horizontal ground. the stone has hit the ground after traveling maximum distance from the base of the pillar...show that the distance is u/g√ (u^2 + 2gh) "

My attempt

If the stone is projected at an angle θ with horizontal, then clearly the time taken till the stone reaches the same height (h) on the other side of its motion is 2usinθ/g and the velocity is usinθ. but it has to travel that extra vertical height h to hit the ground. the initial velocity is usinθ...let the time taken to falling that xtra distance h be t..

so h = usinθ*t + (1/2)*g* t^2

2gh = (gt)^2 + 2gtusinθ

(gt)^2 + 2gtusinθ + (usinθ)^2 = (usinθ)^2 + 2gh

(gt + usinθ)^2 = (usinθ)^2 + 2gh

gt = √{(usinθ)^2 + 2gh} - usinθ

t = 1/g[√{(usinθ)^2 + 2gh} - usinθ]


so total time of flight

T = 2usinθ/g + 1/g[√{(usinθ)^2 + 2gh} - usinθ]

= 1/g [√{(usinθ)^2 + 2gh} +usinθ]

so the horizontal distance traveled

ucosθ *T

= ucosθ* 1/g [√{(usinθ)^2 + 2gh} +usinθ]

= u^2*cosθsinθ/g + ucosθ*√{(usinθ)^2 + 2gh}/g

= u^2* sin2θ/2g + √{(u2sinθcosθ)^2 + 8gh}/2g

= u^2* sin2θ/2g + √{(usin2θ)^2 + 8gh}/2g

Now what to do ?