# Find the initial velocity of the second ball

• gigi9
In summary, the problem involves finding the initial velocity of a ball thrown downward and dropped from different heights, given that both balls reach the ground at the same time. The equations used are h(t)= -9.8t^2+ 19.6 for the dropped ball and h(t)= -9.8t^2+ ct+ 79.6 for the thrown ball. To solve, set both equations equal to 0 and solve for t to find the time it takes for each ball to reach the ground. Then, plug this value of t into the second equation and solve for c to find the initial velocity of the thrown ball.
gigi9

A ball is dropped out of a window 19.6m above the ground. At the same time another ball is thrown straight down from a window 79.6m above the ground. If both balls reach the ground at the same moment, find the initial velocity of the second ball.
***position = s(t)= -16t^2 + ct +d

v(t) = s'(t) = -32t + c

Since they reach the ground, v(final) = 0, you find v(t) = 0.

-32t + c = 0.

t = c/32 s.

The time at which the ball reaches 0 m/s is at c/32 s. It takes the ball c/32 s to reach it's highest point before start falling back to earth. Since the time which the ball reaches the ground is equal to the time it takes to reach the highest point, the answer should be c/32 s.

the velocity before it hits the ground is v(c/32) = -ct + c = c(c - t).

v(t) = s'(t) = -32t + c

Since they reach the ground, v(final) = 0, you find v(t) = 0.

-32t + c = 0.

t = c/32 s.

What? The velocity of a dropped object or one thrown downward is NOT 0 when it hits the ground! Of course, you change that later. What you are doing is finding the highest point (for an object thrown UPWARD) to find the initial velocity of the object. Assuming that "hitting the ground" means coming back to its initial height, that would be the same as the final speed. Normally you do better than this, Prudens Optimus. Did you misread the problem?

NONE of those are true here! One object is dropped (initial velocity 0) and the other is thrown downward (initial velocity unknown- call it c as gigi9 did.
The object dropped has height h(t) at time t given by h(t)= -9.8t2+ 19.6 (height is measured in meters from the ground, t in seconds).
The object thrown downward has height h(t)= -9.8t2+ ct+ 79.6.

Solve -9.8t2+ 19.6= 0 (h= 0 when the object hits the ground) for t, then put that value of t into -9.8t2+ ct+ 79.6= 0 and solve for c (it will be negative, of course).

## 1. What is the formula for finding the initial velocity of the second ball?

The formula for finding the initial velocity of the second ball is v2 = v1 + at, where v2 is the final velocity of the second ball, v1 is the initial velocity of the first ball, a is the acceleration, and t is the time.

## 2. How is the initial velocity of the second ball related to the initial velocity of the first ball?

The initial velocity of the second ball is related to the initial velocity of the first ball through the equation v2 = v1 + at. This means that the initial velocity of the second ball is dependent on the initial velocity of the first ball as well as the acceleration and time.

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

Yes, the initial velocity of the second ball can be negative. This would indicate that the second ball is moving in the opposite direction of the first ball, with a velocity that is decreasing over time.

## 4. What units should be used for the initial velocity of the first ball and the acceleration?

The initial velocity of the first ball and the acceleration should have the same units in order for the equation to work. It is important to use consistent units such as meters per second (m/s) or kilometers per hour (km/h).

## 5. How can the initial velocity of the second ball be measured or calculated?

The initial velocity of the second ball can be measured using a motion sensor or by calculating it using the equation v2 = v1 + at and known values for the initial velocity of the first ball, acceleration, and time. It can also be determined by analyzing the position and velocity data of the two balls over time using tools such as a motion graph.

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