Question

What is the approximate horizontal velocity at which the boy in the picture attached inside threw the ball?

+5 m/s, +20 m/s, +25 m/s, or +30 m/s

***not sure how to figure this out based on the image? thank you:)

Answer 1

30 m/s i believe

Answer 2

how did you find that though? :/ unsure of how to interpret and use the numbers from the image :(

Answer 3

This more of a Physics problem...
it could be solved with Calculus... but it would be somewhat tricky

Answer 4

yes, it's physics! (i have not been able to add subjects lately for some reason!)

Answer 5

seems the most ideal out of the 4 answers if you think about it i mean it wouldnt be the first 2.

Answer 6

I ain't good at physics, but I can try.

Answer 7

okay:)

Answer 8

thank you:)

Answer 9

To do this using Physics, we would need to list 5 variables in both the horizontal and vertical directions the 5 variables are: initial velocity, final velocity, distance, time, and acceleration. Of those variables, which ones do we know in the horizontal direction?

Answer 10

we know time right?

Answer 11

i have the equation x=vx t ? but not sure if that is applicable here?

Answer 12

Notice the trajectory of the ball. When the ball leaves the boy's hand, it has only a
horizontal component of velocity. The only thing that cause the ball to move vertically
down is gravity.
First, find how long it takes for an object to fall from 5 m to the ground in free fall.
That means with an initial vertical velocity of zero.
Then use that time and a distance of 25 m to find the horizontal speed.velocity.

Answer 13

ohh okay:) so i'm multiplying time*25? what would the time be? :/

Answer 14

The distance equation for free fall is:

\(d = \dfrac{1}{2}gt^2\)

You know d. g is a constant. Solve for t.

Answer 15

okay:)

so 25=1/2gt^2 ? like that?

Answer 16

Yes. g is the acceleration due to gravity.
Remember to use the correct value for the units you are working with.

Answer 17

okay:)

so 50=gt^2 ? is that right so far? :/

Answer 18

or do i enter a number for g? :/

Answer 19

Yes, correct so far. g = 9.81 m/s^2

Answer 20

ohh okay:)

and so 50/9.81 = t^2 right?

5.0968399=t^2 ?

Answer 21

t=2.2576181
t is about equal to 2.26 ? :/ did i do that right?

Answer 22

Yes.

Answer 23

yay! :) so what happens from here now?

Answer 24

Wait. I am sorry, but you made a mistake above that I didn't catch till now.
Remember that for now we are only dealing with the vertical displacement.
d in our equation so far is 5 m not 25 m.

Answer 25

This equation you wrote: 25=1/2gt^2
really should be: 5=1/2gt^2

Answer 26

ohhh okay

so it should be 5=1/2gt^2 ?
10=gt^2
10/9.81=t^2
1.0193679=t^2
t=1.0096375

so t is about equal to 1.01 ?

Answer 27

The vertical distance is 5 m, not 25 m.
We will use 25 m soon, but not yet.

Answer 28

ahh okie:)

Answer 29

Great.
The time it took for the ball to fall 5 m is 1.01 s.

Answer 30

yay!

Answer 31

Now we have a ball that moved horizontally 25 m in 1.01 s.
What speed is 25 m in 1.01 seconds?

Answer 32

so it'll be 24.75 right? so about 25?

Answer 33

Correct.

Answer 34

ahh yay!! thank you so much! :)

Answer 35

You're welcome.