How do I solve this problem without knowing the time?

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In summary: Therefore, we can use the first equation to solve for time (t):0=50+9.8t-50=9.8tt=5.1 secondsIn summary, to answer the given questions, we can use the equations d=1/2gt^2 and v=u+at to solve for the maximum height reached by the ball and the time it takes to reach that height. We can also use the equation v=u+at to find the acceleration of the ball during this time.
  • #1
Player7900
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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:

Homework Statement



"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?"


Homework Equations



d = 1/2gt2


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...
 
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  • #2
Your initial vertical velocity is 50m/s
 
  • #3
rock.freak667 said:
Your initial vertical velocity is 50m/s

Right, that is given in the question. I'm not sure what to do with that...
 
  • #4
Player7900 said:
Right, that is given in the question. I'm not sure what to do with that...

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?
 
  • #5
rock.freak667 said:
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?

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)
 
  • #6
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
 
  • #7
Edgar92 said:
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

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?
 
  • #8
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
 
  • #9
1/2mv^2=mgh
work from there
 
  • #10
Edgar92 said:
1/2mv^2=mgh
work from there

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?
 
  • #11
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
 
  • #12
Be sure to post when your done so I know when to stop lol
 
  • #13
Edgar92 said:
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

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...
 
  • #14
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
 
  • #15
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?
 
  • #16
Player7900 said:
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?

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.
 

1. What is the force that causes a ball to move upward when thrown?

The force that causes a ball to move upward when thrown is the initial force applied by the thrower's hand. This force is a result of the muscles in the hand and arm pushing the ball forward.

2. How does gravity affect a ball thrown upward?

Gravity affects a ball thrown upward by constantly pulling it back towards the ground. As the ball moves upward, the force of gravity gradually slows it down until it eventually reaches the peak of its trajectory and starts to fall back to the ground.

3. Why does a thrown ball eventually come back down to the ground?

A thrown ball eventually comes back down to the ground due to the force of gravity. Gravity is a force that attracts objects towards each other, and on Earth, it is what keeps us grounded. As the ball moves upward, the force of gravity pulls it back towards the ground, causing it to fall back down.

4. What factors can affect the height a ball reaches when thrown upward?

The height a ball reaches when thrown upward can be affected by several factors, including the initial force applied by the thrower, the angle at which the ball is thrown, and air resistance. The mass and size of the ball can also play a role in its trajectory.

5. Why do some objects, like a basketball, bounce when thrown upward?

Some objects, like a basketball, bounce when thrown upward because of their elasticity. When an elastic object, such as a basketball, hits the ground, it temporarily deforms and then quickly returns to its original shape, causing it to bounce back up into the air. This bouncing motion continues until the force of gravity eventually overcomes the object's elasticity and it comes to a stop.

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