# Lost on Projectile motion problem

• exk4
In summary, a marble is thrown horizontally from the top of a building at a speed of 12.0 m/s. When it hits the ground, it has a velocity that makes an angle of 30.3 ° with the horizontal. The question is asking for the height from which the marble was thrown, assuming no air resistance.
exk4

## Homework Statement

A marble is thrown horizontally with a speed of 12.0 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 30.3 ° with the horizontal. From what height above the ground was the marble thrown?

-im lost at where to begin

## The Attempt at a Solution

I think the key in this question is whether or not you are ignoring air resistance?

Jared

yes, while ignoring the air resistance

If there is no air resistance, there is nothing acting to decelerate the ball in the horizontal plane. The balls horizontal velocity is a constant 12m/s.

To solve this problem, we can use the equations of motion for projectile motion. The first step would be to identify the known and unknown variables. In this case, the known variables are the initial velocity (12.0 m/s) and the angle of the velocity vector at impact (30.3°). The unknown variable is the height from which the marble was thrown.

Next, we can use the equations of motion to find the time of flight and the horizontal distance traveled by the marble. Then, we can use the vertical displacement equation to find the initial height of the marble.

Time of flight (t) = [2 * initial velocity * sin(angle)] / acceleration due to gravity

Horizontal distance (x) = initial velocity * cos(angle) * t

Vertical displacement (y) = initial height + initial velocity * sin(angle) * t - 0.5 * acceleration due to gravity * t^2

Since the marble was thrown horizontally, the initial height (y) would be zero. Therefore, we can rearrange the vertical displacement equation to solve for the initial height (y).

Initial height (y) = 0 + initial velocity * sin(angle) * t - 0.5 * acceleration due to gravity * t^2

Substituting the values from the given information, we get:

Initial height (y) = 0 + (12.0 m/s) * sin(30.3°) * t - 0.5 * (9.8 m/s^2) * t^2

Solving for t, we get:

t = 1.24 seconds

Substituting this value for t in the horizontal distance equation, we get:

Horizontal distance (x) = (12.0 m/s) * cos(30.3°) * (1.24 seconds)

Horizontal distance (x) = 10.44 meters

Therefore, the initial height of the marble can be calculated as:

Initial height (y) = 0 + (12.0 m/s) * sin(30.3°) * (1.24 seconds) - 0.5 * (9.8 m/s^2) * (1.24 seconds)^2

Initial height (y) = 3.82 meters

Therefore, the marble was thrown from a height of 3.82 meters above the ground.

## What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and then moves under the influence of gravity alone. This type of motion is characterized by a curved path known as a parabola.

## What factors affect projectile motion?

The factors that affect projectile motion are the initial velocity, the angle of launch, and the force of gravity. These factors can influence the distance, height, and time of flight of the projectile.

## How do you calculate the trajectory of a projectile?

The trajectory of a projectile can be calculated using the equations of motion for horizontal and vertical motion. These equations take into account the initial velocity, angle of launch, and acceleration due to gravity.

## What are the applications of projectile motion?

Projectile motion has many real-world applications, such as in sports like baseball and golf, fireworks displays, and even space travel. Understanding projectile motion can also help in designing and predicting the flight paths of missiles and other projectiles.

## How can you solve a lost on projectile motion problem?

To solve a lost on projectile motion problem, you can use the equations of motion and plug in the given values to calculate the unknown quantities. It is also helpful to draw a diagram and break the motion into horizontal and vertical components to better understand the problem.

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