Throwing a ball upward

Player7900

Hello everyone. I'm sorry for asking such basic questions without offering much in a solution attempt, but the problem is - I have no idea how to answer them! I'm obviously a non-scientist and my math skills aren't so great either. But I would like if someone could simply show me HOW to answer these questions, so I could find the answers myself:

1. The problem statement, all variables and given/known data

"A ball is thrown straight upward at a velocity of 50 m/s:
(1) How long will it take for the ball to stop rising?
(2) How far up has the ball traveled in this time?
(3) What is the ball's acceleration during this time?"


2. Relevant equations

d = 1/2gt²


3. The attempt at a solution
I know how to answer #2 if the ball is in freefall, but I don't know what to do if the ball starts at an upward velocity. Ditto for #1 and #3...

rock.freak667

Your initial vertical velocity is 50m/s

Player7900

Right, that is given in the question. I'm not sure what to do with that...

rock.freak667

In your kinematic equations you have

v=u+at
v²=u²+2ad
d=ut+1/2at²

u= initial velocity.

At the highest point, what is the final vertical velocity equal to?

Player7900

0 m/s --- So what would I do with that number? (If I should know by now, again I must apologize as I haven't used anything but basic math in years)

Edgar92

The only two equations you need are
1/2mv^2 kinetic
and
mgh gravitational
if the ball has reached it's max height then there is no kinetic so if you make these equations equal to each other, the masses will cancel out, you no your speed v, you no gravity g, all you need to account for is height h, which is simple rearranging

Player7900

I don't think mass needs to be accounted for, for this question. The question assumes there is no air resistance, and I guess the only relevant factors in figuring out the answers are initial velocity and gravity.

For the first formula I get: 1/2(50)^2 = 1250 J

So, what can I do with that number? Should it help me figure out the height?

Edgar92

Set that formula equal to the gravatational potential equation which is mgh but it's like I said, m is irrelevant and u do no g, all you need is h

Edgar92

1/2mv^2=mgh
work from there

Player7900

1250=gh
1250=(9.8)(h)
1250/9.8=h
h=127.5m

Is that correct? If so, how then would I get the time to reach maximum height?

Edgar92

I don't have a calculator but the equation is right
height is equal to distance
use the equation d=v*t since you know 2 of the 3 variables

t=d/v try it and you should be done

Edgar92

Be sure to post when your done so I know when to stop lol

Player7900

127.5/50=t
t=2.55

The last question is "What is the ball's acceleration during this time?" - What should I do to figure that out?

Also, is there a way to figure out (t) before figuring out (d)? I ask because that question that asks for (t) comes first...

Edgar92

a=v/t
now that you have t and have v aswell, just sub in the appropriate values and you are done. I'm done for the day but that should be enough to finish the rest of this question

Player7900

If anyone else could offer one last bit of advice:

Is there a formula that I can use to find (t) before finding (h), if all that is known is the initial velocity?

rock.freak667

The kinematic equations

v=u+at
v²=u²+2ad
d=ut+1/2at²

u= initial velocity.
v=final velocity

At the highest point, the final vertical velocity is zero.