Resultant Velocity of a Projectile

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Homework Help Overview

The discussion revolves around a projectile motion problem involving an archer shooting an arrow horizontally from a cliff. The problem includes calculating the time of flight, horizontal distance to the target, and the resultant velocity at impact along with the angle with the vertical. The context is set within the framework of kinematics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the assumptions necessary for solving the problem, such as neglecting air resistance and considering uniform gravity. There is also exploration of how to calculate the resultant velocity and angle using vector components.

Discussion Status

Some participants have provided guidance on the calculations, particularly regarding the horizontal and vertical components of velocity. There is ongoing exploration of the assumptions and their implications on the calculations, with no explicit consensus reached on the final values or methods.

Contextual Notes

Participants note the assumption of negligible air resistance and uniform gravitational acceleration. There is also mention of potential ambiguities in the calculations presented, prompting further inquiry into the resultant velocity at different time intervals.

malti001
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Homework Statement



An archer stationed at the edge of a cliff shoots an arrow horizontally at 90ms^-1. The arrow falls down and hits a target. Assuming that the arrow is shot from a height of 20m, calculate:

i. How long it takes the arrow to reach the target. (done)
ii. How far horizontally the target is from the base of the cliff. (done)
iii. The resultant velocity at impact and the angle this makes with the vertical. Mention one assumption made.

Homework Equations



SUVAT equations:

436357594271ec28379aa9b0e6342b5a.png


The Attempt at a Solution



Parts i. and ii. were basically tackled by splitting the problem in half via resolving components vertically and horizontally. Time taken for the arrow was found to be 2s and the target was found to be 180m away from the base of the cliff.

What confuses me in part iii. is the assumption I have to make. I know that resultant velocity is the resultant vector from the horizontal and vertical components, and that the angle is found by resolving the vector diagram.

Am I missing something obvious? Cheers.
 
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You're doing it right bro..
Maybe by the 'assumptions' they mean some basic assumptions like neglecting resistance by air, uniform gravity, etc.
 
Yeah, negligible air resistance is an assumption. (my physics course doesn't feature air resistance at all) But how is this going to help me in finding the resultant velocity at impact?
 
Well there's no horizontal acceleration, so you can take the final horizontal velocity to be 90 m/s.

You can take the final vertical velocity to be ##\vec{a} Δt##.

The resultant velocity would then be : ##\vec{v}_R = \sqrt{ (v_{F_V})^2 +(v_{F_H})^2 }##

Then to get the direction, make your triangle.
 
I think I get it now. The only acceleration present is the downward acceleration due to gravity so obviously the horizontal acceleration would be 0.

Since impact occurs when t=2s, I made a triangle with 90m/s as the horizontal component and 19.6m/s as the vertical component. This is since after 1s, it's accelerating at 9.81m/s per second and thus after 2s, it becomes 19.6m/s (per second). Resultant Velocity was then worked out to be 92.1m/s and the angle that it makes with the vertical was found out to be 77.7°.

This is better summed up in this diagram (it has different numbers though, but still relevant)

u3l2c1.gif


Does any of the above make sense? Lastly, cheers for helping out guys :)
 
Last edited:
malti001 said:
This is since after 1s, it's accelerating at 9.81m/s per second and thus after 2s, it becomes 19.6m/s (per second).

There seems a small ambiguity in that sentence.
Can you find the resultant velocity and angle with vertical after 1.3 sec?
Just to check if you really got it! :smile:
 

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