# Solving a Physics Mystery: Initial Speed & Height of Ball

• Bottomsouth
In summary, the ball was initially traveling at a speed of $v$ and was caught by the outfielder at a height of $h$.
Bottomsouth

## Homework Statement

A batted baseball leaves the bat at an angle of 33.0 above the horizontal and is caught by an outfielder 400 ft from home plate at the same height from which it left the bat

What was the initial speed of the ball?
How high does the ball rise above the point where it struck the bat?

## Homework Equations

the 3 acceleration equations

v = vo + at
v^2 = vo^2 + 2a(deltax)
x = xo + vot + 1/2at^2

## The Attempt at a Solution

I am at a lost to find velocity. Cannot seem to even start the problem.

You will need to split the problem up into horizontal and vertical components of motion. Once you have done that it should be relatively easy to find the initial speed of the baseball. All you have to do is find an equation with variables you know and one that you want.

I believe I will be using components of initial velocity. Still alittle lost on how to proceed.

U0y = U0sin(33)
U0x = U0cos(33)

Ok so you have the components of the initial velocity there. Now you're told that the ball is caught 400ft away at the same height it left the bat. You know there is no acceleration in the horizontal component so which equation can you use to solve for the initial velocity?

Vx=x/t then 0=Vyit = 1/2gt^2

Vcos33*t = ?
Vsin33 - (1/2)gt = 0

I think I am in the right track, still alittle lost.

How far does it travel in the horizontal direction? It tells you in the question. Now you need to eliminate the t variable by rearranging the vertical equation and plug it into the horizontal equation and solve for V.

Goes 400 ft in the horizontal direction.

t=cos33/v

Vsin33 - (.5) -9.8 * cos33/v = 0

Am I getting warmer?

Nearly. $v\cos(33) t = ?$, what distance does the horizontal equation equal? Plus you need to be careful with units. The distance is given in feet and the gravitational constant you're using is in metres per second squared. Decide which units you're going to use.

would it vcos(33)t= 121.92?

400ft = 121.92 m

Bottomsouth said:
would it vcos(33)t= 121.92?

400ft = 121.92 m

Yes

ok haha, what is the 121.92 m? besides the horizontal direction?

Nothing else, that's it.

Physics question!

AHHHHHH I am in serious need of help with my physics problems. I have this assignment due by 11pm tonight, and I am absolutely lost. This stuff will not "click". I was wondering if anyone would be able to help me.

## What is the initial speed of the ball?

The initial speed of the ball is the velocity at which it is launched or released. This can be calculated using the formula v = √(2gh), where v is the initial speed, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the ball is released.

## What is the height from which the ball was released?

The height from which the ball was released is the vertical distance between the point of release and the ground. This can be determined by measuring the height of the launch point or by using the formula h = (v^2)/(2g), where h is the height, v is the initial speed, and g is the acceleration due to gravity.

## How does the initial speed affect the trajectory of the ball?

The initial speed of the ball affects the height and distance it will travel. A higher initial speed will result in a higher trajectory and longer distance traveled, while a lower initial speed will result in a lower trajectory and shorter distance traveled. This is due to the relationship between initial speed and the ball's velocity and acceleration.

## What factors can impact the accuracy of calculating the initial speed and height of the ball?

Some factors that can impact the accuracy of calculating the initial speed and height of the ball include air resistance, friction, and measurement errors. Air resistance can slow down the ball and affect its trajectory, while friction can also alter the ball's movement. Additionally, any errors in measuring the initial speed or height can result in inaccurate calculations.

## Why is it important to accurately determine the initial speed and height of the ball?

Determining the initial speed and height of the ball is important because it allows for a better understanding of the laws of physics and the principles that govern the motion of objects. It can also be useful in various real-life applications, such as sports, engineering, and ballistics. Additionally, accurate calculations can help in predicting and analyzing the behavior of objects in motion.

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