Finding Velocity from only distance

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the performance of a bow that shoots arrows at a 45-degree angle over a distance of 182.8 meters. The original poster is attempting to calculate the initial velocity required for this distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers the horizontal and vertical distances equal due to the angle of launch and attempts to use kinematic equations to find time and velocity. Some participants question the validity of this assumption and the application of the equations, noting the differences between horizontal constant velocity and vertical constant acceleration.

Discussion Status

Participants are exploring various approaches to the problem, including the need for separate equations for horizontal and vertical motion. Some guidance has been offered regarding the use of the "Range" formula and the necessity of simultaneous equations, but no consensus has been reached on a specific method or solution.

Contextual Notes

The discussion is constrained by the original poster's grade level and the complexity of the problem, as well as the requirement to ignore air resistance in the calculations.

Jonathan.rls
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Mod note: Moved from technical section, so no template
Question:
The ace Edward bow can shoot 182.8 meters at a 45 degree angle, it can shoot 12 arrow per minute.

What is the performance of the
Bow?

Attempt:

Basically considering the Dx=182.8 meters I assume the Dy is the same considering its at a 45 degree angle

I tried using (Time)t= sqr root of 2xDy divide by a (-9.81) to find time but when I use the result to find the velocity sand input the velocity into d=ViT+1/2at^2 the distance does not amount to 182.8 meters which I assume it is supposed to

I am in grade 11 taking physics 20 and I need help
@Jaonathan.rls, in future posts of similar homework or homework-like questions, please post them in the Homework & Coursework sections.
 
Last edited by a moderator:
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At 45 degrees, the x and y components of the velocity at the start of flight are the same. So far so good.

The x part of the speed is presumably constant, ignoring air friction. Not sure how valid that is for a 182.8 meter flight. But never mind. You need to ignore friction for the up-and-down part also, or you can't use (Time)t= sqr root of 2xDy divide by a (-9.81).

However, the vertical distance isn't the same as the horizontal distance. The vertical distance is subject to a constant acceleration. The horizontal is at a constant speed. So you can't use the horizontal distance for Dy.

But you don't need it. What you really want is the time as a function of the initial velocity. That will let you get the x distance as a function of initial velocity. Which will then allow you to solve for the initial velocity, since you know the distance.
 
You need to write two equations, one for the vertical motion and one for the horizontal. You should then have two simultaneous equations which can be solved. You can't do it by solving one first. The two equations will be different because one is constant velocity and the other is constant acceleration.
 
You can use the "Range" formula to find the speed, otherwise you have use the
kinematic equations used to derive the formula.
 

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