Lost on Projectile motion problem

[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

Homework Equations

The Attempt at a Solution

 

[JaredJames]

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

Jared

 

[exk4]

yes, while ignoring the air resistance

 

[JaredJames]

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.

 

[rwisz]

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.