# How Long Does a Horizontally Shot Ball Stay Airborne?

• IneedHelp:-)
In summary, the problem involves finding the time and horizontal distance traveled by a ball that is shot horizontally at a speed of 10 m/s from a height of 2 meters. Using the V2 equation, the displacement can be calculated by setting the final displacement to 0 and solving for time. The initial displacement is 2 meters and the vertical displacement is 0 at the top of the trajectory. The time calculated can then be used to find the horizontal distance using the equation s=s_o + ut.
IneedHelp:-)

## Homework Statement

A ball is shot horizontally at a speed of 10 m/s from a height of 2 meters. How long is the ball in the air and how far horizontally does it travel before it hits the ground.

V2 Equation

## The Attempt at a Solution

I am assuming I need to find both the X and Y components right? I know that the final Y velocity must be 0 because it stops at the top of its trajectory. What is throwing me off is the 2 meters it starts at as well as how I am going to get the horizontal distance.

IneedHelp:-) said:

## Homework Equations

V2 Equation

$$s=s_o +ut-\frac{1}{2}gt^2$$

where s=displacement and s0= initial displacement.

So considering vertically, what is s0? and you want to find the 't' for which s=0.

I would approach this problem using the principles of projectile motion. First, I would break the initial velocity of the ball into its x and y components. Since the ball is shot horizontally, the initial velocity in the x-direction is 10 m/s and the initial velocity in the y-direction is 0 m/s.

Next, I would use the equation for the time of flight of a projectile in the y-direction: t = 2vy/g, where t is time, vy is the initial velocity in the y-direction, and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get t = 2(0)/9.8 = 0 seconds. This means that the ball is in the air for 0 seconds before it hits the ground.

To find the horizontal distance traveled by the ball, I would use the equation x = vxt, where x is the horizontal distance, vx is the initial velocity in the x-direction, and t is the time of flight. Plugging in the values, we get x = (10 m/s)(0 seconds) = 0 meters. This means that the ball travels 0 meters horizontally before hitting the ground.

In conclusion, the ball is in the air for 0 seconds and travels 0 meters horizontally before hitting the ground. This is because the initial velocity in the y-direction is 0, meaning the ball does not have any vertical motion and falls straight down.

## What is a projectile shot horizontally?

A projectile shot horizontally is an object that is launched or thrown through the air horizontally, meaning it is parallel to the ground. This type of motion is also known as horizontal motion or motion in the x-direction.

## What are the key factors that affect the motion of a projectile shot horizontally?

The key factors that affect the motion of a projectile shot horizontally are the initial velocity, the angle of launch, and the presence of external forces such as air resistance and gravity.

## What is the formula for calculating the horizontal distance traveled by a projectile?

The formula for calculating the horizontal distance traveled by a projectile is d = v0 * t, where d is the distance, v0 is the initial velocity, and t is the time of flight.

## How does the initial velocity of a projectile affect its horizontal range?

The initial velocity of a projectile has a direct impact on its horizontal range. The greater the initial velocity, the longer the projectile will travel horizontally before hitting the ground.

## Can a projectile shot horizontally ever reach its initial height again?

No, a projectile shot horizontally will not reach its initial height again unless it is launched from a higher elevation or has some external force acting on it to change its trajectory. This is because the force of gravity constantly pulls the projectile downward, causing it to decrease in height over time.

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