Question

5. If a baseball is projected upward from ground level with an initial velocity of 32 feet per second.
a. What is the maximum height reached by the ball?
b. How long is the ball airborne?

Answer 1

what is the question? full question?

Answer 2

5. If a baseball is projected upward from ground level with an initial velocity of 32 feet per second.
a. What is the maximum height reached by the ball?
b. How long is the ball airborne?

Answer 3

Can somebody please help me? :)

Answer 4

you will have to use equations of motion for this, have you read them?

Answer 5

Is the equation -.5gt^2+Vot+So?

Answer 6

Yes, that's the right equation.

What's V₀? S₀? Acceleration?

Answer 7

there are mainly three of them.

Answer 8

final velocity will be zero because you are throwing upwards.

Answer 9

I believe Vo is initial velocity and So is initial starting point, but I am confused how to use the equation to solve

Answer 10

acceleration will be equal to -g

Answer 11

you know the value of g, right?

Answer 12

I do not know the value of g, but I know that v is 32

Answer 13

acceleration due to gravity, g = 9.8m/s^2

Answer 14

V_0 would be the initial velocity (the velocity the ball was thrown up with.
S_0 would be 0 (the problem says we're on level ground)
Acceleration due to gravity is -32 ft/s²

Answer 15

it says we started from ground!!!

Answer 16

then throw the ball from ground upwards
because we are throwing upwards g which is acceleration due to gravity will be taken as negative

Answer 17

Will g always be 9.8 in all velocity type problems? or will it sometimes be -16^2

Answer 18

no, its a constant for only the problems involving acceleration due to gravity

Answer 19

ok

Answer 20

g is 9.8 m/s² AND 32 ft/s². Different units.
If the problem is in feet, use 32. If it's in meters, use 9.8/

Answer 21

because g is acceleration, its unit is m/s^2 and unit of velocity is m/s

Answer 22

Ichiro, read the problem. We're not using metric! We're given feet, so we use feet.

So now we have an equation:
S(t) = 0 + 32t - 16t²

Can you find the two times where S(t) = 0?

Answer 23

2aS = final velocity ^2 - initial velocity ^2

Answer 24

we dont use only 1 equation. and SI unit of length is meter so will have to convert first

Answer 25

So the equation is -.5(9.8)t^2+32t+0 right?

Answer 26

We don't HAVE to use SI. It's good practice for actual science work, but we don't need to over complicate things...

@monkeygirl8, you need to either use 32 ft/s² as g, or convert the initial velocity to m/s

Answer 27

and we dont have to convert the feet in to meter here. initial velocity is 32feet per sec and final velocity is 0

Answer 28

So the equation is -.5(9.8)t^2+32t+0 right?

Answer 29

No.

Answer 30

nope convert the g or velocity

Answer 31

write velocity in m/s or acceleration in feet/s^2

Answer 32

Could someone please show me what the equation would look like, please :)

Answer 33

wait a sec

Answer 34

S(t) = 0 + 32t - 16t² (this is in feet)
OR
S(t) = 0 + 9.7536t - 4.9t² (this is in meters)

Remember, the problem is given in feet, so we need an answer in feet.

Answer 35

I would reccomend:
S(t) = 0 + 32t - 16t²

Answer 36

so I do not need to use g=9.8 anywhere?

Answer 37

No. That's only when the problem is given to you in meters.

Answer 38

what would be the answer then, ddcamp?

Answer 39

The position is given by the equation:
S(t) = c + bt + at²
S(t) = 0 + 32t - 16t²

We can find the maximum height by solving the equation:
\[t=\frac{ -b }{ 2a } = \frac{ -32 }{ 2(-16) } = 1\]
This gives us the time for the maximum point of the equation.

Plug in 1 for t to find the height of the maximum.

Answer 40

and answer?

Answer 41

with my approx, it should be around 15 or 16 feet

Answer 42

I am lost for what to do with the equation. I have a graphing calculator, can I just graph to find the maximum?

Answer 43

\[2gS= Vf^2-Vi^2\]
where g = 32.15 ft/s^2
Vf = 0
Vi = 32ft/s

Answer 44

\[S=Vi^2/(2g)\]

Answer 45

answer is 15.92ft

Answer 46

So 15.92 is the max height. That makes sense.
what is the amount of time it is airborne?

Answer 47

now for the second part:
\[Vf = V i+ g t\]

Answer 48

@monkeygirl8 What class is this for? Because you would use different methods for a math class and a physics class.

Answer 49

\[Vf = 0\]
\[g = 32.15 ft/s^2\]
\[Vi = 32ft/s\]

Answer 50

I am in pre calculus

Answer 51

OK. So, you're working with quadratics?

Answer 52

yes

Answer 53

OK.
So, we use the quadratic equation S(t) = 0 + 32t - 16t²
(in math classes, we use g=32 ft/s² because that makes calculations easier)

Can you find the zeroes of:
S(t) = 0 + 32t - 16t²

Then S(t)=0, we know the ball is at ground level.

Answer 54

the zeros are 2 and 0

Answer 55

Good. So that means, at 2 seconds, the ball hits the ground again.

Answer 56

Thank you so much! You were so much help :)

Answer 57

Did ichiros process for part (a) make sense? He used a physics formula, but I can explain in a way that uses pre-calc.

Answer 58

I think I understand it now. Thank you so much for your help. :)

Answer 59

hey ddcamp, which grade are you in?

Answer 60

I'm a senior. Taking linear algebra and college physics.