Calculating Initial Speed: Solving for Vi in Projectile Motion

[anglum]

Calculating Initial speed?

Homework Statement

A ball is thrown straight up at ground level passes a height of 80.6m in 5.1s.
The acceleration of gravity is 9.8m/s squared. What was its initial speed? answer in units of m/s

Homework Equations

What equation is the correct one to use to calculate this

The Attempt at a Solution

I tried using the formula tup=viy/g

and i also tried doing it as a table... where i calculated the velocity at 5.1 secs to be 15.804 m/s... by dividing 80.6 by 5.1 seconds...

then by adding( since if i was going ahead in seconds instead of back) 9.8 to that speed at 4.1 secs and gettin 25.604 m/s since
and goin up thru till 0 seconds... to get velocity of 65.784

please help me get thru this... this is my last problem for the weekend

 

[learningphysics]

80.6 is not the maximum height. There's a kinematics equation you can use directly and solve for v1. Think about the equations you have at your disposal.

 

[anglum]

i know 80.6 is not the maximum height but I am guessin i cannto use the subtracting gravity from speed... so there is just one kinematics equation that i can use and plug in the data i have that will help me solve this

 

[learningphysics]

i know 80.6 is not the maximum height but I am guessin i cannto use the subtracting gravity from speed... so there is just one kinematics equation that i can use and plug in the data i have that will help me solve this

What is the equation you can use?

 

[anglum]

im not sure what equation i can use which i would plug in the 5.1 seconds and the 80.6m as well as the acceleration of gravity to get my initial speed... tehre is one that i can put all 3 into?

 

[learningphysics]

im not sure what equation i can use which i would plug in the 5.1 seconds and the 80.6m as well as the acceleration of gravity to get my initial speed... tehre is one that i can put all 3 into?

Yup there is. what are your displacement formulas?

 

[anglum]

what displacement formula let's me take into account gravity

 

[anglum]

so would the formula be

Vf=Vi + at

where i solve for Vf as 80.6/5.1? and plug the gravity into the equation for the value of A?

 

[learningphysics]

so would the formula be

Vf=Vi + at

where i solve for Vf as 80.6/5.1? and plug the gravity into the equation for the value of A?

No it's a formula that has d... displacement

 

[anglum]

would it be... Vf squared = Vi squared + 2 A D

where Vf is equal to 80.6/5.1

A = -9.8m/s

D = 80.6 m

?

 

[learningphysics]

would it be... Vf squared = Vi squared + 2 A D

where Vf is equal to 80.6/5.1

A = -9.8m/s

D = 80.6 m

?

Not that equation. The equation has vi, d, t and a. No vf.

 

[anglum]

damn i thought that was the it...

ok so my only other equation that i think is

D = (Vi t) + 1/2 a t squared

so D would = 80.6

a would equal -9.8

would both the t equal 5.1?

then solve for Vi to get the initial speed?

 

[learningphysics]

damn i thought that was the it...

ok so my only other equation that i think is

D = (Vi t) + 1/2 a t squared

so D would = 80.6

a would equal -9.8

would both the t equal 5.1?

then solve for Vi to get the initial speed?

exactly.

 

[anglum]

ok however... if i solve for that i get the initial velocity to be negative?

that doesn't make sense

 

[anglum]

i get 80.6 = Vi(5.1) + 1/2(-9.8 * 5.1) squared

then i get

80.6 = Vi(5.1) + 1/2(-49.980) squared
80.6=Vi(5.1) + 1/2(2498.00)
80.6=Vi(5.1) + 1249
-1168.400 = Vi(5.1)
-229.098 = Vi

that does not seem right to me

 

[learningphysics]

i get 80.6 = Vi(5.1) + 1/2(-9.8 * 5.1) squared

then i get

80.6 = Vi(5.1) + 1/2(-49.980) squared
80.6=Vi(5.1) + 1/2(2498.00)
80.6=Vi(5.1) + 1249
-1168.400 = Vi(5.1)
-229.098 = Vi

that does not seem right to me

It should be:
$$\frac{1}{2}gt^2$$

not
$$\frac{1}{2}(gt)^2$$

only t is squared. not g.

 

[anglum]

or do i only square the time and nto the acceleration of gravity?

 

[anglum]

o sorry u posted faster that i typed that question


so is it (1/2g)(t squared)?

 

[learningphysics]

o sorry u posted faster that i typed that question


so is it (1/2g)(t squared)?

yup.

 

[anglum]

so would the initial speed be equal to 40.794m/s in this particular problem

 

[learningphysics]

so would the initial speed be equal to 40.794m/s in this particular problem

exactly.

 

[anglum]

thank you very much i appreciate you takin the time to help me

 

[rocomath]

so would the initial speed be equal to 40.794m/s in this particular problem

that's what i got.

$$y=y_0+v_{0y}t+\frac{1}{2}gt^2$$

$$v_{0y}=\frac{y-y_0-\frac{1}{2}gt^2}{t}$$

 

[HallsofIvy]

You said "D = (Vi t) + 1/2 a t squared"

That does NOT give you "80.6 = Vi(5.1) + 1/2(-9.8 * 5.1) squared"

Do you see the difference?
D= v_it+ (1/2)a t²

not "D= v_it+ (1/2)(at)²"!