Resultant Velocity of a Projectile

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SUMMARY

The discussion centers on calculating the resultant velocity of a projectile, specifically an arrow shot horizontally at 90 m/s from a height of 20 m. The time taken for the arrow to reach the target is established as 2 seconds, with the horizontal distance to the target being 180 m. The resultant velocity at impact is calculated to be 92.1 m/s, with an angle of 77.7° with respect to the vertical. Key assumptions include neglecting air resistance and considering uniform gravitational acceleration.

PREREQUISITES
  • Understanding of projectile motion and the SUVAT equations.
  • Knowledge of vector resolution in physics.
  • Familiarity with gravitational acceleration (9.81 m/s²).
  • Ability to calculate resultant vectors from horizontal and vertical components.
NEXT STEPS
  • Study the derivation and application of the SUVAT equations in projectile motion.
  • Learn how to resolve vectors and calculate resultant velocities in two-dimensional motion.
  • Explore the effects of air resistance on projectile motion and how to account for it in calculations.
  • Practice calculating angles of impact and resultant velocities for various projectile scenarios.
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to resultant velocity and vector resolution.

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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