Joza said:Yes I agree with that. I have the time it takes to hit the ground from s=ut+(1/2)at^2.
The first rock is in the air for this time plus the time interval when the 2nd rock started. Correct?
Joza said:Ok, I got it!
I realized that distance above the cliff is negligible because it's displacement is 0 back at the cliff when it comes down...
Constant acceleration is the rate of change of velocity that remains the same over a period of time. It is represented by the symbol "a" and is measured in units of meters per second squared (m/s^2).
The constant acceleration can be calculated by dividing the change in velocity (Δv) by the change in time (Δt). This can be represented by the formula a = Δv/Δt.
Constant acceleration is when the velocity of an object changes at a constant rate, while uniform motion is when the velocity remains constant. In other words, in constant acceleration, the object is speeding up or slowing down, while in uniform motion, the object is moving at a constant speed.
To solve a constant acceleration problem, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time. You can also use the formula x = ut + 1/2at^2, where x is the distance traveled.
Some real-life examples of constant acceleration include a car accelerating from a stop, a skydiver falling towards the ground, and a rocket taking off into space. In all of these cases, the velocity of the object is changing at a constant rate over time.