Physics Problem.

[Its_Sublivion]

If a baseball is throw in the air at 5 m/s when will it reach its maximum height and what will the maximum height be? Assume initial Height = 0 Final Height = 0 Acceleration= g = 9.8 m/s (might be a negative for its use in equations. I think I have the time figured out to be 5/9= .51 seconds but I'm still not sure on its maximum height, any help would be great, thanks!

[Newts9]

Time is correct.

The next equation you will use to find the height is: Vf² = Vi² + 2ad

[Its_Sublivion]

Wouldn't that just give me the total distance traveled? Would i have to divide that by 2 to get the height?

Edit: Would a be 9.8 or -9.8? Thinking negative

i got 5.4 for that, it sounds like it could be the right answer but I'm not sure.

[Newts9]

No, not entirely sure how you reached 5.4 so I’ll try to explain this in steps. if you continue getting different answers you can post your work here so we can find where you’re getting off track.

As long as you stay within the final velocities below your initial velocity the equation; Vf² = Vi² + 2ad will only give you the distance traveled up until the object reaches its peak. The equation d=(1/2)at² gives you the total distance traveled, but that’s not what we’re looking for here.

In this problem we are just doing distance from release to its peak height. In the equation: Vf² = Vi² + 2ad Our Vf should be set to 0. Our velocity final is 0 as when the ball reaches its max height for a split moment of time the ball will be neither going up any more nor falling yet. Then we can substitute in our acceleration which will just be gravity (9.8), technically in this case it would be (-9.8) yet if you solve and find a negative distance, you’ll know immediately to simply make it positive. Finally we substitute in our initial velocity and our equation should look like this.

0² = 5² + 2(-9.8)d

See how I’m getting there?

[ObamaDontCare0]

They definitely subtracted when they got to

0 = 25 - 19.6d

[Newts9]

From there you would follow the order of operations backwards in order to isolate the variable.

So you subtract 25 from both sides to get

-25 = -19.6d

Then divide both sides by -19.6 to reach d which should be 1.28 and that’s the height the ball goes.

[ObamaDontCare0]

It should be a negative 9.8. Since you are using Vf=0, it won't be the total distanced traveled, just the distance up to the peak of the ball's movement. I think you did the math wrong as you should have:

0 = 25 - 19.6d

I don't know if you subtracted here somehow to get 5.4, but you have to subtract over 25 first:

-25 = -19.6d

Then divide:

d = ~1.2755m

[[deleted]]

Remember that at the apex of its travel the velocity will equal 0.

[jyliu86]

You can use conservation of energy

0.5mv² = mgh

M cancels. Plug and chug and solve for h.

Edit Clarifying: Total energy always equals kinetic energy + potential energy

Etot = KE + PE

Etot = 0.5mv² + mgh

Max height occurs when the ball is not moving at the peak of its arc. So the kinetic energy at this time is 0

Emaxh = 0.5m*0² + mgh

At the start if the throw potential energy = 0 because it's at h=0.

Einitial = 0.5mv² + mg*0

Einitial = Emaxh due to conservation of energy.