Falling Particle: Solving for Initial Velocity and Acceleration

  • Thread starter Thread starter sudec
  • Start date Start date
  • Tags Tags
    Falling Particle
AI Thread Summary
The discussion focuses on solving for the initial velocity (v0) and acceleration (a) of a particle thrown upwards, which passes a distance "d" twice—once while ascending and once while descending. The correct formulas derived for these calculations are v0 = d * [(t1 + t2) / (t1 * t2)] and a = (2 * d) / (t1 * t2). The problem emphasizes starting with the position equation for an object under constant acceleration. Participants express difficulty in arriving at the correct solution, highlighting the need for a clear understanding of the motion equations. The discussion ultimately clarifies the relationship between time, distance, and velocity in projectile motion.
sudec
Messages
2
Reaction score
0

Homework Statement



Particle was thrown directly upwards. Point which is in distance “d” from the beginning of movement is passed by the particle two times. Once going upwards in time “t1” and second time going downwards in time “t2” from the beginning of the movement. What was the initial speed(v0) and the acceleration(a) of the particle?

Homework Equations



I really do not know how to start with this example.

The Attempt at a Solution



I have tried every possible solution that I was capable of, but I’m not able to come up with the correct answer which should be

v0=d*[(t1+t2)/(t1*t2)]

and

a=(2*d)/(t1*t2)
 
Physics news on Phys.org
Start with the equation for position o an object under constant acceleration.

Point which is in distance “d” from the beginning of movement is passed by the particle two times.

This defines the coordinate system for you: initial velocity is v0 and initial height is 0.
 
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