Question

A tennis ball is struck and departs from the racket of Prof. Jacome horizontally with a speed of 30m/s. The ball hits the court at a horizontal distance of 20 m from the racket. How far above the court is the tennis ball when it leaves the racket?

Answer 1

The tennis ball is travelling horizontally with a speed of 30m/s, and ends up 20m away from the racket. This is all we are given.

Answer 2

So, we will assume the racket is \(y_0\) from the ground.

Answer 3

\(y_0{~}-{\rm the~initial~hieght~of~the~ball}\).
\(v_{x0}=v_x=30~{\rm m/s}\) (\(v_{x0}=v_x\) because air resistance is neglected)
\( x=20\) (distance before landing)

Answer 4

Can you work out the time it takes the ball to travel 20m horizontally?

Answer 5

(You velocity in the horizontal is unchanged, because we neglect air resistance)

Answer 6

For time, can we use t= v - vo / a

Answer 7

No, you don't need any kinematics for this.

Answer 8

Your velocity in the horizontal is 30m/s (and it doesn't change).
The distance in the horizontal that you traveled is 20m.
So the time is?

Answer 9

(Yes, it is that simple, because the motion in the horizontal and the motion in the vertical are independent of each other.)

Answer 10

\(\displaystyle \frac{\Delta x}{\Delta t}=v_x\) Right?

So, \(\displaystyle \Delta t=\frac{\Delta x}{v_x}=\frac{20{\rm m}}{30{\rm m/s}}=?\)

Answer 11

.666 seconds

Answer 12

yes, 2/3 seconds.

Answer 13

(I will use a fraction. I like to use exact values, and as I find the answer, then approximate .... well, in any case, we want the concept, so if you have any conceptual questions, let me know.)

Answer 14

\(\color{black}{\displaystyle v_{y}(t)=v_{y_0}-gt }\) familiar with this?

Answer 15

hmm is what is vy and vy0 in your formula?

Answer 16

Alternatively, \(\color{black}{\displaystyle v=v_{0}-gt }\),
and this is for the velocity in the y direction, where g is gravity.

Answer 17

I am using a subscript of y to denote that it is a motion or velocity in the vertical.

Answer 18

In other words, you agree that we can model the velocity in the\(y\) at any time \(t\) with \(\color{black}{\displaystyle v(t)=v_{0}-gt }\)

Answer 19

and the consequence of this equation (via integration) is
\(\color{black}{\displaystyle y(t)=(v_{0})t-\frac{1}{2}gt^2+y_0 }\)

Answer 20

This equation models the vertical displacement.
Seen this one before?

Answer 21

hmm well if your talking about integration i am not familiar with it. we're still just learning basic physics

Answer 22

No, the equation, \(\color{black}{\displaystyle y(t)=(v_{0})t-\frac{1}{2}gt^2+y_0 }\)

Answer 23

Did you see this (above) equation?

Answer 24

Is this the first time you see this equation, or is it familiar to you?

Answer 25

i dont think ive seen this equation

Answer 26

Oh, it's one of the kinematics, and this is the appropriate equation to use in this case.

Answer 27

Why?
Because you are given the time it takes to fall, t=2/3 seconds,
you are given y(t)=0 at this time (b/c at this time ball hits the ground).
Also you know gravity is 9.8m/s^2, and you know that the initial velocity in the vertical (the \(v_0\)) is 0.

Answer 28

Is this formula the same as y= V0yt+ 1/2 ayt ?

Answer 29

yes, it is the same formula.
\(\color{black}{\displaystyle y=v_{0y}+\frac{1}{2}a_yt}\)

Answer 30

Do you know all of the information to plug in, or should we work on this together?

Answer 31

(You are certainly welcome to try it, and I will check you .... wanna do this?)

Answer 32

Yes! i would like to do so

Answer 33

Sure:)
Go ahead, and take your time.

Answer 34

How would i go about finding Voy?

Answer 35

First, I want to adjust your formula.

Answer 36

\(\color{black}{\displaystyle y(t)=v_{0y}+\frac{1}{2}a_yt+y_0}\)

Answer 37

This is more appropriate, because the initial displacement in the y, is not 0.
And y(t), because this is the height at time t.

Answer 38

As far as the velocity goes, you are given \(v_{0x}\), and \(v_{0y}\) is not mentioned, so we are assuming that initial the ball's speed is perfectly horizontal.

Answer 39

(In other words, your initial velocity in the y, just the instant the ball leave the racket, is equal to 0.)

Answer 40

and knowing this you can just cross
\(\color{black}{\displaystyle y(t)=\color{red}{\bcancel{\cancel{\color{black}{v_{0y}}}}}+\frac{1}{2}a_yt+y_0}\)

Answer 41

So, with a zero initial velocity in the y you will have
\(\color{black}{\displaystyle y(t)=\frac{1}{2}a_yt+y_0}\)

Answer 42

The time it takes to fall is \(t=2/3{~}{\rm s}\), and at this time you know that \(y(t)=0\) (because this is when the ball reaches the ground). We also know \(\color{black}{\displaystyle a_y=g=-9.8~{\rm m/s^2}}\).

Answer 43

So, there you have it all to find the initial displacement in the vertical, \(y_0\).
(This \(y_0\) is the initial height of the ball, and this is the same height at the very instant the ball leaves the racket.)

Answer 44

oh so the Yo would be 20m ?

Answer 45

Please read the question, and tell me what this value (20 m) corresponds to.

Answer 46

You are given everything (I listed) ... all that remains for you is to plug in the information into \(\color{black}{\displaystyle y(t)=\frac{1}{2}a_yt+y_0 }\) and solve for \(y_0\).

Answer 47

Im not sure if i follow you

Answer 48

Its very confusing

Answer 49

Do you understand why it is \(\color{black}{\displaystyle y(t)=\frac{1}{2}a_yt+y_0 }\), (without adding \(v_{0y}t\) to this)?

Answer 50

(Because originally, the equation is \(\color{black}{\displaystyle y(t)=v_{0y}t+\frac{1}{2}a_yt+y_0 }\),
however we got rid of that term by noticing something.
-- What did we notice?)

Answer 51

Voy is not mentioned in the problem

Answer 52

We noticed that \(\color{black}{\displaystyle v_{0t}=0 }\), because the only velocity that is there, is the velocity in the horizontal (and not in vertical).

Answer 53

And since \(v_{0y}=0\) we can just write the same without \(v_{0y}t\).

Answer 54

For this reason, instead of the original equation, I have now this:
\(\color{black}{\displaystyle y(t)=\frac{1}{2}a_yt+y_0 }\)

Answer 55

Can you tell me, what is \(\color{black}{\displaystyle a_y }\) ?

Answer 56

Oooh! ok ok Ay would be the acceleration

Answer 57

Yes, and in this case the acceleration in the vertical is the what?

Answer 58

BTW, COrrection t is squared.
\(\color{black}{\displaystyle y(t)=\frac{1}{2}a_yt^2+y_0 }\)