I can always rely on Maple Story for homework help.

[tHEeVILkItTY]

OKAY HERE
THIS IS TEH RITE 1



Explanation:
"X = VT" is simply "distance = velocity x time"

Y = V1T + 1/2 GT^2 is one of the kinematic equations.
V1 refers to initial velocity, and since we're using this equation for the y-axis, V1 is 0 (he throws it horizontally, so there's no downward velocity on the ball when he throws it).
Thus Y = 1/2 GT^2

So basically you first convert the 60 ft. into meters and the 90 mph into meters/second.
Then, you plug the meter value of the horizontal distance (X) as well as the initial horizontal velocity (V) into X = VT.
You solve for T, and plug in T into Y = 1/2 GT^2.
Since G is a constant (9.8 m/s^2), you can solve for Y, your answer.

The correct answer should be 2.227272727272727272727..........

Oh cool, I spent 100k on a wrong answer.

Ohwell. xD

I told you to use basick maf skillz but nuuuuuuuuuuu you just had to go about it the right way.

Gaaah! D:

[EpitomeofMe]

pshh, you could've just come to Negi and he would've given you the right answer with a diagram for free :p

[Lamiss]

OMG that was soo smartt lollll

[Atlatl]

Is Neji really right? Its 1/12th of a second to reach home, by my common sence, it shouldn't drop by 2 feet, less probably. I don't know how to solve the problem though. So neji might be right




EDIT:!!!!!!!!! Neji, if the ball drops 2 METERS than it's not going to reach home, or at least it'll bounce. 2M is about the height of a man, a fairly tall man too.

EDIT2: I got 1280ft per second and I don't think that it's 9 meters/sec

[NegiWarrior]

Is Neji really right? Its 1/12th of a second to reach home, by my common sence, it shouldn't drop by 2 feet, less probably. I don't know how to solve the problem though. So neji might be right




EDIT:!!!!!!!!! Neji, if the ball drops 2 METERS than it's not going to reach home, or at least it'll bounce. 2M is about the height of a man, a fairly tall man too.

EDIT2: I got 1280ft per second and I don't think that it's 9 meters/sec

Yeah, the logic is messed up.
But the equations and junk should be accurate.

I did the conversions with an online converter, so those are 100% accurate.


EDIT: Wait, CRAP.
I forgot to square the time on the vertical equation.
Everything is right up until the answer, which should be 1.012396694... meters.


90mph is = 90 x 5280 ft/hour
= 90 x 5280/3600 ft/s
= about 0.3 x 90 x 5280/3600 m/s
= 39.6 m/s, but the more accurate answer is 40.2336 m/s

[darksidehero]

That's a pissweak throw.

[tHEeVILkItTY]

pshh, you could've just come to Negi and he would've given you the right answer with a diagram for free :p

I did. He said he forgot everything, so I went and FM spammed.

._.

That's a pissweak throw.

I thought 90mph was fast. >__>

[Thorr]

You don't even need to go too far with this.

The only this you need to know is the time between mound and home. 90mph = 132 ft/sec. The time is thus .454545 seconds.

Gravity is not a velocity, it's an acceleration. You can't multiply it directly in that formula.

Gravity accelerates at 9.8m/sec^2. This equals 32.15223 ft/sec^2. You now have the number of seconds (.454545) so plug it in -

.454545^2. = .2067 secs.

The formula for Distance is s = ut + 1/2 at^2. s = distance, u = initial velocity (0 in this case) t = time (.454545) a = acceleration (31.15223 fps),

Distance = 0 * .454545 + 1/2(31.15233 * .2067)
Distance = 0 + 1/2(6.44)
Distance = 3.21ft.

So the ball should drop 3.21 ft. Aim it at the level of the guys head (6ft) and it should drop into the strike zone (3ft). That's been the rule f thumb as long as I have been pitching.

[NegiWarrior]

You don't even need to go too far with this.

The only this you need to know is the time between mound and home. 90mph = 132 ft/sec. The time is thus .454545 seconds.

Gravity is not a velocity, it's an acceleration. You can't multiply it directly in that formula.

Gravity accelerates at 9.8m/sec^2. This equals 32.15223 ft/sec^2. You now have the number of seconds (.454545) so plug it in -

.454545^2. = .2067 secs.

The formula for Distance is s = ut + 1/2 at^2. s = distance, u = initial velocity (0 in this case) t = time (.454545) a = acceleration (31.15223 fps),

Distance = 0 * .454545 + 1/2(31.15233 * .2067)
Distance = 0 + 1/2(6.44)
Distance = 3.21ft.

So the ball should drop 3.21 ft. Aim it at the level of the guys head (6ft) and it should drop into the strike zone (3ft). That's been the rule f thumb as long as I have been pitching.

You can multiply gravity into the equation because it is an acceleration. o_O

I used the simple X = VT to find out the total time it took the ball to move horizontally, and then used that same time value for the vertical movement.

We used the same exact formula for the latter half of the problem (ie, the vertical distance formula.
Plus, my work was in meters so I didn't have to convert the value of gravity into ft/sec^2.

Our answers are only different by a little bit because you messed up by retyping the value of gravity in ft/sec^2 from 32.15233 into 31.15233. :P

I still think my method is easier!

[Sasuke-Kun]

Your geniusness hurts MUH HEAD.