solving for velocity

[aoneill]

In a football game, David punts the ball from a height of 2 feet. The ball stays in the air for 3.2 seconds before hitting the ground. find the initial upward velocity of the kick. I have been using the formula h=-1/2g(t^2) +vt+h where the second h is the initial height. Supposedly the answer is approx 50.6 ft/sec but when I plug the numbers in where I think they go this is not what I get. Please help as soon as possible. I have a test tomorrow. Thank you.

 

[Deleted member 4993]

In a football game, David punts the ball from a height of 2 feet. The ball stays in the air for 3.2 seconds before hitting the ground. find the initial upward velocity of the kick. I have been using the formula h=-1/2g(t^2) +vt+h where the second h is the initial height. Supposedly the answer is approx 50.6 ft/sec but when I plug the numbers in where I think they go this is not what I get. Please help as soon as possible. I have a test tomorrow. Thank you.

Yep I get 50.6 ft/sec too.

Since you didn't show any work - I can't tell where is the pain!!!!

 

[aoneill]

h=-1/2g(t^2)+vt+h
I solved for v using 0 for the first h, as the ball eventually hit the ground:
0=-1/2 (9.8)(3.2^2)+v(3.2)+2
0=-4.9(10.24)+v(3.2)+2
0= -48.126+3.2v
48.126/3.2 = v
15.055=v

 

[Deleted member 4993]

h=-1/2g(t^2)+vt+h
I solved for v using 0 for the first h, as the ball eventually hit the ground:
0=-1/2 (9.8)(3.2^2)+v(3.2)+2 <<<< That should be 32

0=-4.9(10.24)+v(3.2)+2
0= -48.126+3.2v
48.126/3.2 = v
15.055=v

.

 

[aoneill]

why should h (initial height) be 32?

 

[Deleted member 4993]

why should h (initial height) be 32?

Either you are color-blind - or you are not reading my response carefully.

I am saying g = 32 (ft/s²)

instead of

9.8 (m/s²)

 

[soroban]

Hello, aoneill!

In a football game, David punts the ball from a height of 2 feet.
The ball stays in the air for 3.2 seconds before hitting the ground.
Find the initial upward velocity of the kick.

I have been using the formula:.$\displaystyle y\:=\:h_o + v_ot -\frac{1}{2}gt^2$

where: .$\displaystyle \begin{Bmatrix}y &=& \text{height of ball} \\ h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ g &=& 32\text{ ft/sec}^2 \end{Bmatrix}$

Supposedly the answer is approx 50.6 ft/sec,
but when I plug the numbers in where I think they go, this is not what I get.

We have: .$\displaystyle y \:=\:2 + v_ot - 16t^2$


We are told "The ball stays in the air for 3.2 seconds before hitting the ground."
This means: when $\displaystyle t = 3.2$, the ball is on the ground: .$\displaystyle y =0.$

We have: .$\displaystyle 0 \:=\:2 + v_o(3.2) - 16(3.2)^2 \quad\Rightarrow\quad 3.2v_o \:=\:161.84$

. . . $\displaystyle v_o \:=\:\dfrac{161.84}{3.2} \:=\:50.575$

Therefore: .$\displaystyle v_o \;\approx\;50.6\text{ ft/sec}$

 

[aoneill]

While I appreciate your help, I don't appreciate your tone - you are very rude.