Ball thrown up, find max height - Given one velocity in midflight

In summary, the problem involves finding the maximum height reached by a ball thrown upwards from a certain point near a tall office building. The ball is later seen passing by a person's window with a speed of 2.9 m/s, located 5.1 meters above the starting point. Using kinematic equations and taking into account the acceleration due to gravity to be 9.82 m/s^2, the maximum height is calculated to be a little greater than 12 meters. However, there may have been a mistake in the initial calculation and it is advised to redo the problem. A helpful hint is to temporarily ignore the observer's position and focus on the height the ball reaches with a starting velocity of 2.9 m/s
  • #1
ichivictus
39
0

Homework Statement



A ball is thrown upwards from a point near the sidewalk next to a tall office building. At some later time an occupant of the building, exactly 5.1 meters above the point from which the ball was released, sees it pass her window with a speed of 2.9 m/s.

Calculate the maximum height reached by the ball.

You may neglect air resistance. Take the acceleration due to gravity to be 9.82 m/s^2


Homework Equations


Kinematic Equations

y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]


The Attempt at a Solution



Using Kinematic equation Vf^2=Vo^2+2ay I solve for Vo
(2.9)^2=Vo^2 - 19.64*5.1
Vo = 10.01 m/s

To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo.

0 = 10.01^2 - 19.64y
y=5.1 m

Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.
 
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  • #2
Your value for vo does not look right. I think you forgot to add the (2.9)^2 term.
 
  • #3
nasu said:
Your value for vo does not look right. I think you forgot to add the (2.9)^2 term.

That must have been it. I just redid the problem and received an answer a little greater than 12 meters. Not sure where I went wrong exactly, but no worries.

Thanks :)
 
  • #4
Here's how I would look at it. Start from the window.

The ball is traveling upwards at 2.9 m/s, so how high does it go from there?

Just add that answer to the starting point of 5.1 m.
 
  • #5
ichivictus said:
That must have been it. I just redid the problem and received an answer a little greater than 12 meters. Not sure where I went wrong exactly, but no worries.

Thanks :)
I suggest you redo the problem again. You didn't show your work, so there's no knowing where you made a mistake.

Hint: Temporarily ignore the fact that the observer was 5.1 meters above the ground. Suppose you threw a ball upwards at 2.9 m/s. How high will it fly given g=9.82 m/s2?
 

1. What is the formula for calculating the maximum height of a ball thrown up with a given velocity in midflight?

The formula for calculating the maximum height of a ball thrown up with a given velocity in midflight is:
h = (v2sin2θ)/2g
Where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

2. How do I determine the initial velocity of the ball if I know the maximum height and the angle of projection?

To determine the initial velocity of the ball, you can use the following formula:
v = √(2gh)/sinθ
Where v is the initial velocity, h is the maximum height, θ is the angle of projection, and g is the acceleration due to gravity.

3. Can I use the same formula to find the maximum height if the ball is thrown at an angle other than 90 degrees?

Yes, the same formula can be used to find the maximum height even if the ball is thrown at an angle other than 90 degrees. However, in this case, the angle of projection (θ) needs to be considered when calculating the initial velocity.

4. Is the maximum height the same as the height at which the ball was thrown?

No, the maximum height is the highest point that the ball reaches during its flight. The height at which the ball was thrown may be different, depending on the initial height of the throw.

5. Can the maximum height be greater than the initial height of the throw?

Yes, the maximum height can be greater than the initial height of the throw if the ball is thrown at an angle other than 90 degrees. This is because the initial velocity in the vertical direction is greater than the initial velocity in the horizontal direction, causing the ball to reach a higher point before falling back to the ground.

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