Calculating Time of a Ball Thrown with Velocity 39m/s | No Air Resistance

[physicsgurl12]

Homework Statement

A ball is thrown straight up with a velocity of 39m/s. how much time passes before the ball hits the ground. No air resistance.
4s-wrong
1.2s
2.4s
8s

Homework Equations

v=d/t a=v/t

The Attempt at a Solution

9.8/39= .02
39/9.8=3.97959

 

[berkeman]

Homework Statement

A ball is thrown straight up with a velocity of 39m/s. how much time passes before the ball hits the ground. No air resistance.
4s-wrong
1.2s
2.4s
8s

Homework Equations

v=d/t a=v/t

The Attempt at a Solution

9.8/39= .02
39/9.8=3.97959

It would be better if you started with the general equation for vertical motion (under constant downward acceleration, like due to gravity):

$$y(t) = y_0 + v_{y0} t - \frac{1}{2} g t^2$$

If you start with that equation, you get two solutions for the time when the object is at ground level...

 

[DaveC426913]

If you start with that equation, you get two solutions for the time when the object is at ground level...

Well, I hope at least one of them is at t=0...

 

[berkeman]

Well, I hope at least one of them is at t=0...

 

[physicsgurl12]

It would be better if you started with the general equation for vertical motion (under constant downward acceleration, like due to gravity):

$$y(t) = y_0 + v_{y0} t - \frac{1}{2} g t^2$$

If you start with that equation, you get two solutions for the time when the object is at ground level...

okay thanks for the equation . but i solved it out and got .1 ?

 

[berkeman]

okay thanks for the equation . but i solved it out and got .1 ?

Nope.

Please show your work, and we'll see if we can spot the error...

 

[physicsgurl12]

okay let's see. i guess i did. 0=39m/s*t-4.9t^2
4.9/39=.1256 yeah that's not even real algebra

 

[berkeman]

okay let's see. i guess i did. 0=39m/s*t-4.9t^2
4.9/39=.1256 yeah that's not even real algebra

Almost there.

You should be solving for the time t, not 1/t...

Write out the algebra steps more clearly after that equation, and be sure to solve to for t.

 

[physicsgurl12]

1.2??

 

[berkeman]

1.2??

No.

Your equation here is correct:

0=39m/s*t-4.9t^2

Just solve it for t. You came close when you tried before. Show us each algebra step. There are two terms on the righthand side of the equation. How do you move one of them to the LHS of the equation? Can you then cancel anything out?

Solve for t.

 

[physicsgurl12]

0=39m/s*t-4.9t^2

Just solve it for t. You came close when you tried before. Show us each algebra step. There are two terms on the righthand side of the equation. How do you move one of them to the LHS of the equation? Can you then cancel anything out?

Solve for t.

0= 39m/s*t-4.9*t^2
4.9*t^2=39m/s*t
4.9*t=39m/s
t=7.959?

 

[berkeman]

0= 39m/s*t-4.9*t^2
4.9*t^2=39m/s*t
4.9*t=39m/s
t=7.959?

Bingo!

So which multiple choice answer is correct?

 

[physicsgurl12]

D. 8s yay!