Calculating Time and Velocity of a Ball on a Ramp - Confirming Understanding"

[J_o_S]

can i just get someone to confirm this?

just want to make sure I'm understanding this correctly.


A ball is rolled up a ramp. The initial velocity and the acceleration down the ramp due to gravity is given.

Vi = 10 m/s
a = 2 m/s

-------------------------------------------------
To find out when the ball is at rest I use:

Vf = Vi + at

where,
vf = 0
vi = 10
a = -2

---------------------------------------------------

To find out the when the ball's velocity is 6 m/s I use the same formula:


Vf = Vi + at

where
vf = 6
vi = 10
a = -2
-----------------------------------------------------

to find out how long it takes for the ball to get back to its starting point:

(The time it takes to get to the highest point is already known from the first part.)


I first find d:

d = [(Vi + Vf)/2]t


Then the time it takes to get back down can be found by:

d = Vit + .5at^2

d = .5at^2


Then I add the time it takes to get up with the time it takes to get down for the total time.


And the time it takes to get up should = the time it takes to get down.

--------------------------------------------------------------------


Everything look ok here?

thanks for looking

 

[learningphysics]

Everything looks good to me. But for the last part I'd do it like this:

Solve
d = Vit + .5at^2

where Vi=10, a=-2, d = 0. You'll get two solutions. One is t=0 (that's when the ball starts), and the other solution is the one you're looking for, when the ball comes back down.

The way you did it (dividing into the up trip, and the down trip) is also correct. Try both ways and confirm that you get the same answer.

 

[thepatientmental]

over this and confirming.

Yes, your understanding and calculations seem to be correct. To confirm, the formula Vf = Vi + at is used to calculate the velocity of the ball at a specific time, and the formula d = Vit + .5at^2 is used to calculate the distance traveled by the ball at a specific time. The time it takes for the ball to reach its highest point can be found by dividing the initial velocity by the acceleration due to gravity (t = Vi/a). And as you mentioned, the time it takes for the ball to go up and come back down should be the same, since it follows the same path. Overall, your approach and calculations seem to be accurate. Good job!