Archer on a Cliff: Can Arrow Reach Mark?

  • Thread starter dboy83
  • Start date
In summary: That's why we don't need to include the negative sign in the equation. In summary, we can use the formula 0.5*g*t^2=100 to calculate the time t of the flight of the arrow, which we can then use in the formula v0*t = 35*t to determine the x displacement of the arrow and see if it has hit the target or not. To find the initial velocity needed for the arrow to reach the target at a 0 degree angle, we can use the kinematics equations and solve for v0.
  • #1
dboy83
34
0
An archer stands on a cliff that is 100m high, shooting at a mark on the ground below 430m away from the bottom of the (vertical) cliff. The archer is capable of shooting an arrow with an initial speed of 35 m/s at any angle you choose.

A) If the archer shoots horizontally ( at an angle of zero degrees with respect to the horizontal), will the arrow reach the mark? why? If not, will shooting at any angle above the horizontal help? Why or why not?

B) What initial speed of the arrow is needed to be in order for the arrow to reach the target mark ( still with an angle of zero degrees)?:rofl:
 
Physics news on Phys.org
  • #2
You need to show your work so far. What equations of motion apply to this problem?
 
  • #3
Just some hints, use your kinematics equations, the 4 equations. And the time for vertical motion and horizontal motion is the same.

Hope it helps
 
  • #4
lost...

berkeman said:
You need to show your work so far. What equations of motion apply to this problem?


I don't know which one to use. Where do i start? Please somebody help.
 
  • #5
what formula

dboy83 said:
An archer stands on a cliff that is 100m high, shooting at a mark on the ground below 430m away from the bottom of the (vertical) cliff. The archer is capable of shooting an arrow with an initial speed of 35 m/s at any angle you choose.

A) If the archer shoots horizontally ( at an angle of zero degrees with respect to the horizontal), will the arrow reach the mark? why? If not, will shooting at any angle above the horizontal help? Why or why not?

B) What initial speed of the arrow is needed to be in order for the arrow to reach the target mark ( still with an angle of zero degrees)?:rofl:


does anybody know what formula I am supposed to use?
 
  • #6
What is the equation that relates changes in velocity to the initial velocity and the acceleration? What is the slightly longer equation that relates changes in position to initial position, initial velocity, and acceleration?
 
  • #7
The velocity in the x direction is constant and it equals the initial velocity v0 = 35 m/s. The y direction velocity equals gt. Hence, the path in the x direction equals v0*t = 35*t (1), and in the y direction 0.5*g*t^2. Now if the arrow falls to the ground, the path that it has to cross in the y direction equals 100 m, so 0.5*g*t^2=100. Solving this equation gives us the time t of the flight of the arrow. Putting that time into equation (1) gives us the x displacement of the arrow, so we can tell if it has his the target or not.
 
  • #8
hmm

radou said:
The velocity in the x direction is constant and it equals the initial velocity v0 = 35 m/s. The y direction velocity equals gt. Hence, the path in the x direction equals v0*t = 35*t (1), and in the y direction 0.5*g*t^2. Now if the arrow falls to the ground, the path that it has to cross in the y direction equals 100 m, so 0.5*g*t^2=100. Solving this equation gives us the time t of the flight of the arrow. Putting that time into equation (1) gives us the x displacement of the arrow, so we can tell if it has his the target or not.

You guys make it look so easy... Ok, so 0.5*g*t^2=100

I plug in -9.80 for g and multiply that by 0.5, which gives me -4.9.
so I could rewrite the equation as -4.9*t^2=100 If I try to solve for t I get the square root of a negative number which is not possible. ?
 
  • #9
I'm thinking

dboy83 said:
You guys make it look so easy... Ok, so 0.5*g*t^2=100

I plug in -9.80 for g and multiply that by 0.5, which gives me -4.9.
so I could rewrite the equation as -4.9*t^2=100 If I try to solve for t I get the square root of a negative number which is not possible. ?

is 0.5*g*t^2=100 supposed to be 0.5*g*t^2=-100 ? ( the only difference is the negative sign in the 100.
 
  • #10
dboy83 said:
is 0.5*g*t^2=100 supposed to be 0.5*g*t^2=-100 ? ( the only difference is the negative sign in the 100.

The ball is 'falling down', so gravity is positive.
 
  • #11
radou said:
The ball is 'falling down', so gravity is positive.
Yeah, the acceleration and velocity and displacement are all in the same direction, so everything is positive.
 

1. How does an archer determine the distance required to reach a target on a cliff?

An archer can determine the distance required by using their knowledge of the trajectory of an arrow, the angle of the cliff, and the height of the target. They can also use tools such as range finders or mathematical equations to calculate the distance.

2. What factors affect an archer's ability to reach a target on a cliff?

The main factors that can affect an archer's ability to reach a target on a cliff include the archer's skill and accuracy, the type and quality of their bow and arrows, the wind and weather conditions, and the angle and height of the cliff.

3. Can an arrow really reach a target on a cliff?

Yes, an arrow can reach a target on a cliff if the archer has the necessary skill and equipment, and the conditions are favorable. Archers have been able to hit targets at great distances for centuries, and with modern technology, it is even more achievable.

4. How does the angle of the cliff affect an archer's shot?

The angle of the cliff can greatly affect an archer's shot. If the cliff is steep, the archer may need to aim higher to compensate for the downward trajectory of the arrow. If the cliff is at an angle, the archer may need to adjust their aim to the left or right to account for the arrow's trajectory.

5. Is it more difficult to hit a target on a cliff compared to a flat surface?

Yes, it can be more difficult to hit a target on a cliff compared to a flat surface. The angle and height of the cliff can make it challenging for the archer to accurately judge the distance and trajectory of their shot. However, with practice and proper technique, skilled archers are able to hit targets on cliffs with precision.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
5K
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top