Resultant Velocity of a Projectile

AI Thread Summary
An archer shoots an arrow horizontally at 90 m/s from a height of 20 m, taking 2 seconds to reach the target, which is 180 m away from the cliff base. The resultant velocity at impact is calculated using the horizontal component of 90 m/s and the vertical component of 19.6 m/s, resulting in a total velocity of 92.1 m/s. The angle of impact with the vertical is determined to be 77.7°. Key assumptions include neglecting air resistance and considering uniform gravitational acceleration. The discussion emphasizes the importance of resolving the projectile's motion into horizontal and vertical components to find the resultant velocity and angle.
malti001
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
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:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top