Displacement from angle and initial velocity

[Planefreak]

Homework Statement

A ball is thrown horizontally from the top of a building at 45.6 m/s. We want to know how long it takes for the ball to drop below an angle of 13 degrees. Ignore air resistance.

Homework Equations

I'm assuming the projectile equation is necessary (delta)y = V_yit + .5(a)t² other than that I'm lost

The Attempt at a Solution

I've tried it a few different ways. First thought was to use the information given to write two equations. One being the flight path of the ball and the other of a line at 13 degrees. This gave me an incredibly small amount of time and was incorrect.
The two lines would be -4.9t² and -tan(13)x They both should intersect twice. Once being at zero the other the point we are interested in. This didn't work out.

I tried rewriting equations to setup a different system of equations but ultimately couldn't get anything out of it.

Somehow I got .55 and thought that was right but according to the teacher it is not. So what do I need to do to solve the problem?

 

[heth]

Any time an angle is mentioned, look for a right-angled triangle and components.

Try drawing a sketch of what the projectile will look like at the exact moment in time you're after, spot a triangle, and label it with horizontal and vertical displacement components, plus any other information you know.

It doesn't matter if you don't know values for the displacements yet - just use letters to represent the triangle's sides.

Once you've done that, try to write down some equations for the horizontal and vertical displacements involved and see how that works out.

 

[baba89]

I would approach this problem by first identifying the relevant equations and principles that apply to the situation. In this case, we can use the projectile motion equation, which describes the motion of an object in a constant gravitational field without air resistance. This equation is:

y = y0 + v0y*t + (1/2)gt^2

where y is the vertical displacement, y0 is the initial vertical position, v0y is the initial vertical velocity, t is time, and g is the acceleration due to gravity (9.8 m/s^2).

We know that the ball is thrown horizontally, so the initial vertical velocity (v0y) is zero. We also know that the initial vertical position (y0) is the height of the building. Therefore, we can rewrite the equation as:

y = h - (1/2)gt^2

where h is the height of the building. We want to find the time it takes for the ball to drop below an angle of 13 degrees, which means that the vertical displacement (y) is equal to the height of the building multiplied by the tangent of 13 degrees. Therefore, we can set up an equation:

h*tan(13) = h - (1/2)gt^2

This is a quadratic equation in terms of t, which can be solved using the quadratic formula. Once we solve for t, we can determine the time it takes for the ball to drop below 13 degrees.

It's important to note that this solution assumes no air resistance, which may not be a realistic assumption. In a real-world situation, air resistance would affect the trajectory of the ball and the time it takes to drop below an angle of 13 degrees. Therefore, the calculated time may not be entirely accurate, but it can provide a good estimate.