# Max height of thrown object with 12m/s at 1/5 of height

• brundlfly
In summary: You must know the maximum height in order to find the initial speed. In summary, the conversation discusses a problem where an object is shot vertically upward with a speed of 12 m/s and reaches one fifth of its maximum height. The equations used for free falling objects are mentioned, and the attempt at solving the problem is described. The correct approach is explained, where the maximum height and initial speed must be known in order to find the speed at one fifth of the maximum height.
brundlfly

## Homework Statement

An object is shot vertically upward and has a speed of 12 m/s when it reaches one fifth of its maximum height.
a. What is the initial speed of the object?
b. What is the maximum height of the object?

## Homework Equations

Free falling objects

## The Attempt at a Solution

I used the formula vf^2 = vi^2 - 2g(yf - yi) and got 144 = 0 + 19.6(1/5 - 0), I solved for 1/5 so I have 144/19.6 = 1/5 and got 1/5 = 7.35 meters. This times 5 yields 36.73 meters. This would be it for part b. But I am stuck on part a where I have to find the speed. Any advise on how to approach this? Thank you.

brundlfly said:
I used the formula vf^2 = vi^2 - 2g(yf - yi) and got 144 = 0 + 19.6(1/5 - 0)

You got a negative value (because -2g(yf - yi) is negative), and then you probably thought "oops, the square of velocity can't be negative, so let's change the sign and hope for the best". It does not work that way.

You should have asked yourself "why did I get a negative value where only a positive one is possible? Where did I go wrong"?

voko said:
You got a negative value (because -2g(yf - yi) is negative), and then you probably thought "oops, the square of velocity can't be negative, so let's change the sign and hope for the best". It does not work that way.

You should have asked yourself "why did I get a negative value where only a positive one is possible? Where did I go wrong"?

Not really, the reason why I ended up with a positive value instead of a negative is because

144 = 0 - 2(-9.6)(1/5 height) = 144 = 0 -(-19.6)(1/5 height) = 144 = 19.6 (1/5 height). If you don't agree let me know.

brundlfly said:
Not really, the reason why I ended up with a positive value instead of a negative is because

144 = 0 - 2(-9.6)(1/5 height) = 144 = 0 -(-19.6)(1/5 height) = 144 = 19.6 (1/5 height). If you don't agree let me know.

The basic kinematic equation is ##v_f^2 = v_i^2 + 2 a(s_f - s_i) ##. Adapted to free fall, it is either ## v_f^2 = v_i^2 + 2g (y_f - y_i) ##, where ## g ## is then taken negative, or ## v_f^2 = v_i^2 - 2g (y_f - y_i) ##, where ## g ## is then taken positive. But you cannot have both the equation in the form ## v_f^2 = v_i^2 - 2g (y_f - y_i) ## and ## g ## negative in it.

brundlfly said:
I used the formula vf^2 = vi^2 - 2g(yf - yi) and got 144 = 0 + 19.6(1/5 - 0), I solved for 1/5 so I have 144/19.6 = 1/5 and got 1/5 = 7.35 meters. This times 5 yields 36.73 meters. This would be it for part b. But I am stuck on part a where I have to find the speed. Any advise on how to approach this? Thank you.

yf is not 1/5.
When the speed is zero, the object has reached the maximum height. (h_max).
When the speed is 12 m/s the object height is 1/5 from this maximum height.

1/5 is not a height in meters (or other unit of length) but a fraction.
Saying that 1/5=7.35 m is nonsense.

## 1. What is the equation for calculating the maximum height of a thrown object with a velocity of 12m/s at 1/5 of the height?

The equation for calculating the maximum height of a thrown object is h = (v^2 * sin^2θ)/(2g), where v is the initial velocity, θ is the angle of the throw, and g is the acceleration due to gravity (9.8m/s^2).

## 2. How do you determine the angle at which the object was thrown?

To determine the angle of the throw, you can use the equation θ = sin^-1(h * 2g/v^2), where h is the maximum height, g is the acceleration due to gravity, and v is the initial velocity. Note that this equation assumes that the object was thrown from ground level.

## 3. What is the maximum height of the thrown object if the initial velocity is changed to 15m/s?

To calculate the new maximum height, you can use the equation h = (v^2 * sin^2θ)/(2g), where v is the new initial velocity (15m/s), θ is the angle of the throw, and g is the acceleration due to gravity (9.8m/s^2).

## 4. Can the maximum height of the thrown object be greater than the initial height?

Yes, it is possible for the maximum height of the thrown object to be greater than the initial height. This can occur if the object is thrown at an angle greater than 45 degrees, as the vertical component of the initial velocity will be greater than the initial height.

## 5. How does the maximum height of the thrown object change if the initial velocity is doubled?

If the initial velocity is doubled, the maximum height of the thrown object will quadruple. This is because the maximum height is directly proportional to the square of the initial velocity.

• Introductory Physics Homework Help
Replies
9
Views
972
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
25
Views
456
• Introductory Physics Homework Help
Replies
9
Views
5K
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
897