Question

A good fast pitcher can throw a ball horizontally with a speed of 30 m/s. if home plate is 18 m from the pitchers mound, how far will the ball drop in that distance?

Answer 1

So first you want to find out how long it takes the ball to get to the home plate.

Given

v = 30 m/s
d = 18m

and there's no acceleration in the x direction,, right?

How long would it take the ball to travel that distance?

Answer 2

how far will the ball drop in that distance

Answer 3

Right. You know that it will drop a certain distance due to gravity, right?

Answer 4

yeah im just not sure which equations to use to find the exact amount the ball dropped

Answer 5

That's why I asked about time. Knowing the time it took for the ball to go the distance is the time you can plug in to your equation of acceleration to find how for it went

Answer 6

the ball wnt 18 m.. so i have to find the acceleration? and thats distance divided by time right?

Answer 7

the speed is 30 m/s

Answer 8

Not quite (the velocity is 30 m/s in the positive x direction). The velocity is distance/time; the acceleration is velocity/time

You know what the acceleration is, though! - It's gravity, going straight down in the y direction. There is no acceleration in the x direction.

Answer 9

you're right about the speed, the "not quite" was about the comment before that.

Answer 10

so what equation would i use to find out how far it dropped in 18m?

Answer 11

sup dominique this guy is awesome

Answer 12

Remember that distance = velocity / time

\[ v = d/t\]

so

\[t = d /v\]

And to find how far something moved while being accelerated is



What's the vi in

Answer 13

GAH, equation editor is driving me crazy

d = vi^2 + 1/2 at^2

Answer 14

\[d = v_{i}^2 + \frac{1}{2} at^2\]

What's the vi in that equation?

Answer 15

yeah im colpletely lost

Answer 16

@dwade this guy is pretty awesome haha

Answer 17

ok ok ok i see now

Answer 18

:D

Answer 19

so the time is 0.6

Answer 20

what would be the a?

Answer 21

yah. The a is just good ol' gravity. -9.8 m/s^2 (assuming we're on earth that is)

Answer 22

and what about gravity?

Answer 23

oh okay

Answer 24

gravity is the only acceleration in the problem. If there weren't gravity, the ball would keep going past home plate forever at 30 m/s

Answer 25

~wasn't

Answer 26

so is vi 3.19?

Answer 27

Not quite. That's just a general equation for any given distance, velocity, and acc. The vi is the initial velocity in the direction we're looking at. So for us it looks like this

h = (Viy)^2 + (1/2) gt^2



I should have made the d different, my bad.

Answer 28

does that make sense?

Answer 29

and h would be the answer?

Answer 30

h would be the answer. Do you see why?

Answer 31

not really

Answer 32

and what do i plug in for viy?

Answer 33

Viy
\[v_{iy}\] is the initial velocity in the y direction - how fast it's moving down when we first start out. Is the ball moving down at all when the guy first throws it?

Answer 34

no so its 0 right?

Answer 35

so then h= -1.7658?

Answer 36

yah!! and that's what I got.

Answer 37

dude youre so awesome

Answer 38

and that's -1.7658 m Always remember the units! ^^

Answer 39

thank you so much

Answer 40

Love me some physics :) Glad I could help :)

Answer 41

now i have 2 more questions. could you help me out with that?

Answer 42

fo sho

Answer 43

Fuzzball, the cat, slides off a table while chasing a red dot from a laser. if the table is 0.82 m tall and the cats horizontal velocity was 2.3 m/s, how far away from the table does the cat land on the floor?

Answer 44

I swore there was an open question about a cat named fuzzball but I can't find it. Was that you?

So this is almost exactly opposite the process from last time.


(a should be ay=g
(ax = 0)


\[h = d_y = v_{iy} t + \frac{1}{2}a_yt^2\]

\[L = d_x = v_{ix}t + \frac{1}{2}a_xt\]