# Solving for Initial Velocity & Max Height of Thrown Ball

• quicksilver123
In summary, the conversation discusses the problem of finding the initial velocity and maximum height of a ball thrown upwards, with a given velocity at 3/5 of its maximum height. The formula vf^2 = vi^2 -2g(yf-yi) is provided, with the direction of up being specified as positive. The conversation includes attempts and concerns about units and negative values, ultimately resulting in the successful solution of the problem.
quicksilver123

## Homework Statement

A ball is thrown upwards. Its velocity at 3/5 its max height is 16.0m/s
What is its initial velocity?
What is its max height?

## Homework Equations

vf^2 = vi^2 -2g(yf-yi)

define up as positive

## The Attempt at a Solution

I ran through numerous attempts on paper but can't seem to find a proper answer. The closest I got was an answer that spit out a negative height... help? Somewhat urgent as I need to bring this question in tomorrow for homework.

Thank you![/B]

Show your attempt, wrong or right.

Sigh. I crossed it all out on my paper, otherwise I'd just take a picture and scan it.

0=(16m/s)^2 -2(-9.8m/s/s)(2h/5)
-(16m/s)^2=18.6m/s/s(2h/5)
-(16m/s)^2= (7.44h)m/s/s
-256m^2/s^2 = h(7.44)m/s/s

Two issues here being that
a) the units do not match since I haven't multiplied in height, so operations may be difficult
b) there is a negative, so the height will come out as a negative, which seems illogical as I have specified up as positive

Last edited:
quicksilver123 said:
Sigh. I crossed it all out on my paper, otherwise I'd just take a picture and scan it.

0=(16m/s)^2 -2(9.8m/s/s)(2h/5)
-(16m/s)^2=-18.6m/s/s(2h/5)
Why didn't you cancel the minus signs on both sides here?
-(16m/s)^2= (7.44h)m/s/s
-256m^2/s^2 = h(7.44)m/s/s

Two issues here being that
a) the units do not match since I haven't multiplied in height, so operations may be difficult
Looks to me like m/s2 on the right will cancel out the same on the left leaving m. Thus the units of h will be m, which is correct.
b) there is a negative, so the height will come out as a negative, which seems illogical as I have specified up as positive
Nope. You just didn't cancel the signs at a previous step.

No, that was a typo. I fixed it. Stated in the original post that up was to be positive. Therefore the acceleration is negative.
Thank you for your comment on the units but the height still appears to turn out negative.

quicksilver123 said:
No, that was a typo. I fixed it. Stated in the original post that up was to be positive. Therefore the acceleration is negative.
Thank you for your comment on the units but the height still appears to turn out negative.
Then you've introduced an excess negative by both subtracting the term 2gh and making the constant g negative.

You can either treat g as a positive constant and account for the direction in the algebra, or you can make "2ad" a positive term and let g be negative to account for the direction.

The general form of the equation is:

##v_f^2 - v_i^2 = 2 a d##

so then:

##-v_i^2 = -v_f^2 + 2 a d##
##v_i^2 = v_f^2 - 2 a d##

and if ##v_f = 0## and ##a = g## is a negative value,

##v_i^2 = -2 g d##

will yield a positive result for d.

quicksilver123
Thank you!
I have a result for height. However, attempting to find the initial velocity... my result is the same as the given velocity (when the ball is a fraction of the max height).

I used the same equation and substituted height.

quicksilver123 said:
Thank you!
I have a result for height. However, attempting to find the initial velocity... my result is the same as the given velocity (when the ball is a fraction of the max height).

I used the same equation and substituted height.

Sorry, it was my mistake. I solved the question. thank you very much for your help.

Okay! Cheers.

## 1. What is the formula for calculating initial velocity of a thrown ball?

The formula for calculating initial velocity, or the velocity at which the ball is thrown, is: v0 = √(g * h), where g is the acceleration due to gravity (9.8 m/s2 on Earth) and h is the maximum height reached by the ball.

## 2. How do I measure the maximum height reached by a thrown ball?

The maximum height reached by a thrown ball can be measured using a measuring tape or ruler. Simply measure the vertical distance from the ground to the highest point the ball reaches in its trajectory.

## 3. Can the initial velocity of a thrown ball be negative?

Yes, the initial velocity of a thrown ball can be negative if the ball is thrown downwards, towards the ground. In this case, the initial velocity is represented by a negative number in the formula.

## 4. Is air resistance taken into account when solving for initial velocity and max height of a thrown ball?

No, the formula for calculating initial velocity and max height assumes that there is no air resistance. In reality, air resistance can affect the trajectory of a thrown ball and should be taken into account for more accurate calculations.

## 5. Can I use the same formula to solve for initial velocity and max height of a ball thrown on a planet with a different gravitational acceleration?

No, the formula for calculating initial velocity and max height is specific to Earth's gravitational acceleration. For planets with different gravitational accelerations, the formula would need to be adjusted accordingly.

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