Initial Velocity & Maximum Height??

Hi kjackson73,

Vf = Vi + at  Where Vf = final velocity, Vi = initial velocity, a = acceleration due to gravity, t = time in seconds

I'm looking at when throwing the ball to when the ball at its maximum height. Then the time would be 1.45/2 = .725, because 1.45 would be the time it takes for the ball to reach it maximum height + the time it takes for the ball to fall down.

Vf of the ball would be 0 m/s at its maximum height.

0 = Vi + (-9.8) * (.725)      I'm assuming that up is positive
Vi = 7.105 m/s This means that Vi is 7.105 m/s going upward

For the maximum height, there are 2 ways to do it.

1 way:  Vf2 = Vi2 + 2ad
                 0     = (7.105)2 + 2 * (-9.8) * d     I'm still assuming that upward is positive.
    50.48 = 19.6d ====> d = 2.59 m  So the maximum height would be 2.59m

Another way to do this is set the kinetic energy equal to the potential energy
(1/2)mv2 = mgh        the "m" cancels out
.5v2 = 9.8 * h
.5 (7.105)2 = 9.8 * h   ===> 2.57m ~ 2,59m

Initial velocity = 7.105m/s
Maximum height = 2.59m

Hope this helps,
Laser

The time to rise to maximum height will be the same time that it takes to fall.

Take your 1.45 seconds, and divide by 2. This gives you 0.725 seconds. This is the time that it takes to rise to its maximum height. It is *also* the time that it takes to fall back down to the baseball player.

You can use this equation which figures out time to maximum height:

Tf = SQRT [ (2H) / g ]

Where Tf is the time to rise (0.725 seconds), H is the maximum height, and g is Earth gravity (9.80665 m/s^2).

Isolate the H

Tf^2 = (2H) / g

(Tf^2) * g = 2H

[ (Tf^2) * g ] / 2 = H

Now just plug in your known time Tf and gravity

H = [ (0.725 s)^2 * (9.80665 m/s^2) ] / 2

H = [ (0.525625 s^2) * (9.80665 m/s^2) ] / 2

H = 5.155 m

Note that the units cancel out.

This is the answer to "Maximum Height" 5.155 m

Initial velocity is going to be the same as *final* velocity, because that's the way it works. So go to this equation:

Tr = (Voy) / g

Where Tr = time to rise (which is the same as time to fall, as above! 0.725 s) and Voy is initial velocity.

Solve for Voy

Tf * g = Voy

Insert values

Voy= (0.725 s) * (9.80665 m/s^2)

Voy = 7.110 m/s

Which is your answer to Initial Velocity 7.110 m/s

It's true that you can do it in a single equation H = Yo + (Voy * Tr) - (0.5 * g * Tr^2) ... but I prefer breaking it into simple pieces.

Hi ARM0714,


Your equation is right; however, the math is a little bit off. I think that you forgot to divide by 2 after multiplying .525625 and 9.8

Best regards,
Laser

We still derived the same answer, sometimes I miscarry the decimal or round awkwardly!