Projectile Motion & Vectors: Acceleration and Axis Conventions Explained

AI Thread Summary
The discussion focuses on projectile motion, specifically the constant downward acceleration of 10 m/s². Participants confirm that the negative sign in the vertical axis (denoted by j) indicates downward direction. The importance of writing kinematic equations for clarity in solving projectile motion problems is also emphasized. Overall, the key takeaway is the consistent downward acceleration in projectile motion and its representation in vector notation. Understanding these concepts is crucial for solving related physics problems effectively.
Sarah00
Messages
64
Reaction score
1

Homework Statement


screenshot_8.png


Homework Equations


-

The Attempt at a Solution


I know that the acceleration is downward always with value of 10m/s2
The answer is D? because j denote vertical axis and negative sign means downward?
[/B]
 
Physics news on Phys.org
It wouldn't hurt to write down the kinematic equations for the projectile motion in step 2.
 
Sarah00 said:

Homework Statement


screenshot_8.png


Homework Equations


-

The Attempt at a Solution


I know that the acceleration is downward always with value of 10m/s2
The answer is D? because j denote vertical axis and negative sign means downward?[/B]
Yes.
 
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...
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}...
Back
Top