Finding Velocity from only distance

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To find the velocity of the bow shooting arrows at a 45-degree angle, it's essential to recognize that the horizontal and vertical components of motion are governed by different principles. The horizontal distance (182.8 meters) is traveled at a constant speed, while the vertical distance is affected by gravity, requiring separate equations for each component. The vertical motion involves constant acceleration due to gravity, while the horizontal motion assumes constant velocity. To solve for the initial velocity, one must set up simultaneous equations for both the horizontal and vertical motions. Using the "Range" formula or kinematic equations will provide the necessary calculations to determine the bow's performance accurately.
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.
 
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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