How much time elapses before the stone hits the ground?

[D4b34r5]

Homework Statement

A stone is launched straight up by a slingshot. Its initial speed is 20.2 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.

A: How high above the ground does the stone rise?
B: How much time elapses before the stone hits the ground?

Homework Equations

Y=V_it+ 1/2a_yt²
v_f²-v_i²/a_y=T

The Attempt at a Solution

I have tried the first formula with the time from the second and can't get the answer. I have tried everything I can think of and was even working with others at school to try and get this one. I have seen the other problem identical to this listed but I couldn't understand it. Please Help

 

[LowlyPion]

Homework Statement

A stone is launched straight up by a slingshot. Its initial speed is 20.2 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.

Aow high above the ground does the stone rise?
Bow much time elapses before the stone hits the ground?

Homework Equations

Y=V_it+ 1/2a_yt²
v_f²-v_i²/a_y=T

The Attempt at a Solution

I have tried the first formula with the time from the second and can't get the answer. I have tried everything I can think of and was even working with others at school to try and get this one. I have seen the other problem identical to this listed but I couldn't understand it. Please Help

Welcome to PF.

I'm sure you meant this equation to be:
(v_f²-v_i²)/2a_y=Y

Maybe that will be all you need?

ay = -g
Y + 1.4 = Ymax

 

[D4b34r5]

Thanks for the welcome. And yes I had that, I just input the equation wrong into the computer.

When I go through and try an answer I have gotten 2.06 for the time it takes for it to stop rising which I would think i could just input into the formula to have it be some thing like y=1.40m+20.2m/s(2.06s)+1/2(-9.8m/s²)(2.06s)². I end up getting 22 but I don't think that is the answer, it just doesn't make sense with it rising for 2 seconds with an initial velocity of 20m/s...

 

[LowlyPion]

Thanks for the welcome. And yes I had that, I just input the equation wrong into the computer.

When I go through and try an answer I have gotten 2.06 for the time it takes for it to stop rising which I would think i could just input into the formula to have it be some thing like y=1.40m+20.2m/s(2.06s)+1/2(-9.8m/s²)(2.06s)². I end up getting 22 but I don't think that is the answer, it just doesn't make sense with it rising for 2 seconds with an initial velocity of 20m/s...

Use the equation I suggested first.

That's (0² - (20.3)²)/2(-9.8) = Y
where your final velocity is 0.
Then to that number you add the additional 1.4 m.

Now use the height Y to figure Time to MAX and then use Y + 1.4 to figure time to fall.. These two times you can calculate simply with Y = 1/2 g t²

Then add the two time results together. That's Total time.

 

[D4b34r5]

Thank you! Finally got it.