Would anybody be able to explain this problem and its equations for me? I don't understand it at ALL.
1)A ball is thrown upwards from a 100 m high cliff at 7.5 m/sec. If you ignore air resistance
A) What is its velocity after 7.0 sec?
B) How long until it is at its high point?
C) Where is it (Vertical height related to the ground) after 7.0 sec
D) When does it reach the ground?
E) What is the total distance moved?
F) WHat is the ball's speed just before hitting the ground?
2) A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s^2. What is the speed of the car after it has traveled 55m. Answer in meters per second.
Please, I really need somebody to explain this to me ASAP. I don't necessarily need the answers, but I NEED the equations! I just don't understand! And none of these are trick questions.
B) How long until it is at its high point?
at the top of the ball's trajectory
C) Where is it (Vertical height related to the ground) after 7.0 sec
use the equation with
D) When does it reach the ground?
set , solve for t
E) What is the total distance moved?
2(displacement up) + 100
F) WHat is the ball's speed just before hitting the ground?
|v| when it hits the ground (use the solution for part D)
go to the link and learn about motion in one dimension ...
http://www.physicsclassroom.com/class/1DKin/U1L6a.cfm
Well i found this site is very useful defined the theory
Motion in One Dimension
We can take the third kinematics equations given by skeeter and modify it a bit to meet our needs. We are told the car begins at rest ( ) and accelerates uniformly at and undergoes a displacement of , so the equation becomes:
Now just plug in the given data to find the magnitude of the final velocity vector, which is the speed.
Suppose we did not have the formula handy. We can find it by reasoning as follows:
We know the distance traveled is the average velocity times the time traveled:
We know uniform acceleration is the change in velocity divided by the change in time:
We have 2 expression for time t, so equate them to get:
Cross-multiply: