Freely falling bodies

In summary, a ball is thrown straight upward and reaches a maximum height of 18 m. The question asks at what height above its launch point the speed of the ball decreases to one-half of its initial value. The relevant equations are vi=0, d=18m, and g=9.81.
  • #1
mandaa123
8
0

Homework Statement



A ball is thrown straight upward and rises to a maximum height of 18 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

Homework Equations


i don't know.



The Attempt at a Solution


vi=0, d=18m, g=9.81, this is all i know.
 
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  • #2
mandaa123 said:

Homework Statement



A ball is thrown straight upward and rises to a maximum height of 18 m above its launch point. At what height above its launch point has the speed of the ball decreased to one-half of its initial value?

Homework Equations


i don't know.



The Attempt at a Solution


vi=0, d=18m, g=9.81, this is all i know.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2
 
  • #3


I would approach this problem by first identifying the relevant equations and principles that govern the motion of freely falling bodies. From the given information, we can use the equation for the displacement of an object under constant acceleration: d = vi*t + (1/2)*a*t^2, where d is the displacement, vi is the initial velocity, a is the acceleration, and t is the time.

In this case, we know that the initial velocity, vi, is 0 since the ball is thrown straight upward. We also know that the acceleration, a, is equal to the acceleration due to gravity, which is approximately 9.81 m/s^2. We are trying to find the height, d, at which the speed of the ball is one-half of its initial value, so we can set vi/2 as the final velocity, vf, in the equation. Therefore, we have:

d = vi*t + (1/2)*a*t^2
d = (vi/2)*t + (1/2)*a*t^2
d = (1/2)*a*t^2

To find the time, t, at which the ball reaches this height, we can use the equation for the final velocity of an object under constant acceleration: vf = vi + a*t. Since we know that vf = vi/2 and vi = 0, we can solve for t:

vf = vi + a*t
vi/2 = 0 + a*t
t = vi/(2*a)

Plugging this value for t into the equation for displacement, we get:

d = (1/2)*a*(vi/(2*a))^2
d = (1/2)*a*(vi^2/(4*a^2))
d = (1/8)*vi^2/a

Therefore, at a height of (1/8)*vi^2/a, the speed of the ball will be one-half of its initial value. Plugging in the given values, we get:

d = (1/8)*(0 m/s)^2/(9.81 m/s^2)
d = 0 m

This means that the speed of the ball decreases to one-half of its initial value at the same height from which it was launched, which makes sense since gravity causes the ball to slow down as it rises and then speeds it up as it falls back down. I hope this helps to
 

1. What is a freely falling body?

A freely falling body is an object that is falling under the influence of gravity alone, with no other external forces acting on it.

2. What is the acceleration of a freely falling body?

The acceleration of a freely falling body is approximately 9.8 meters per second squared, or 9.8 m/s^2. This is also known as the acceleration due to gravity.

3. How does the mass of a freely falling body affect its acceleration?

The mass of a freely falling body does not affect its acceleration. All objects, regardless of their mass, will fall with the same acceleration due to gravity.

4. How does air resistance affect a freely falling body?

Air resistance, also known as drag, can slow down the acceleration of a freely falling body. This is because as the body falls, it encounters air particles which push against it, creating a force that opposes gravity.

5. What is the formula for calculating the distance traveled by a freely falling body?

The formula for calculating the distance traveled by a freely falling body is d = 1/2 * gt^2, where d is the distance, g is the acceleration due to gravity, and t is the time the body has been falling for.

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