How Far Will the Stone Travel Horizontally Before Hitting the Ground?

[professordad]

Homework Statement

A stone is thrown upward from the top of a building at an angle of 30.0 degrees to the horizontal and with an initial speed of 20.0 m/s. The height of the building is 45.0 m.

Where does the stone strike the ground?

Homework Equations

v_{x0} = v_0\cos{\theta_0} where v_{x0} is the original velocity x component, v_0 is the original velocity, and \theta_0 is the original angle (not sure if we use this)

Also v_{y0} is the original velocity y component.

The Attempt at a Solution

I'm not sure how to proceed here. We're finding horizontal distance, so no equations concerning v_{y0} would work, and v_{x0} = v_0\cos{\theta_0} only works if the vertical distance is the same as the original vertical distance..

 

[Sleepy_time]

Hi professordad. If you're confident in using vectors and integrating, for this problem you can use \mathbf{F}=m\mathbf{\ddot{r}}=m(0,-g,0), separate the components and integrate along with the boundary conditions \mathbf{x}_0=(0,h,0), \mathbf{v}_0=(v_0cos(\theta),v_0sin(\theta),0). After obtaining an equation for x(t) and y(t), substitute y=0 (when the y-axis meets the ground) and solve for t, and use this value in x(t).

 

[berkeman]

Homework Statement

A stone is thrown upward from the top of a building at an angle of 30.0 degrees to the horizontal and with an initial speed of 20.0 m/s. The height of the building is 45.0 m.

Where does the stone strike the ground?

Homework Equations

v_{x0} = v_0\cos{\theta_0} where v_{x0} is the original velocity x component, v_0 is the original velocity, and \theta_0 is the original angle (not sure if we use this)

Also v_{y0} is the original velocity y component.

The Attempt at a Solution

I'm not sure how to proceed here. We're finding horizontal distance, so no equations concerning v_{y0} would work, and v_{x0} = v_0\cos{\theta_0} only works if the vertical distance is the same as the original vertical distance..

Hi professordad. If you're confident in using vectors and integrating, for this problem you can use \mathbf{F}=m\mathbf{\ddot{r}}=m(0,-g,0), separate the components and integrate along with the boundary conditions \mathbf{x}_0=(0,h,0), \mathbf{v}_0=(v_0cos(\theta),v_0sin(\theta),0). After obtaining an equation for x(t) and y(t), substitute y=0 (when the y-axis meets the ground) and solve for t, and use this value in x(t).

Doesn't seem like you need to use vectors or integration for this problem, unless I'm missing something.

Just write the two equations for the motion in the vertical and horizontal directions. The horizontal velocity stays constant (ignoring air resistance). The vertical motion follows the equations for motion given a constant acceleration (gravity).

You use the initial vertical velocity and position to figure out how long it takes the ball to hit the ground. Then use that time and the horizontal velocity to figure out how far out it hits.

Lets see some equations!

 

[Sleepy_time]

Doesn't seem like you need to use vectors or integration for this problem, unless I'm missing something.

Just write the two equations for the motion in the vertical and horizontal directions. The horizontal velocity stays constant (ignoring air resistance). The vertical motion follows the equations for motion given a constant acceleration (gravity).

You use the initial vertical velocity and position to figure out how long it takes the ball to hit the ground. Then use that time and the horizontal velocity to figure out how far out it hits.

Lets see some equations!

No not required, just a preference

 

[azizlwl]

Homework Statement

A stone is thrown upward from the top of a building at an angle of 30.0 degrees to the horizontal and with an initial speed of 20.0 m/s. The height of the building is 45.0 m.

Where does the stone strike the ground?

Homework Equations

v_{x0} = v_0\cos{\theta_0} where v_{x0} is the original velocity x component, v_0 is the original velocity, and \theta_0 is the original angle (not sure if we use this)

Also v_{y0} is the original velocity y component.

The Attempt at a Solution

I'm not sure how to proceed here. We're finding horizontal distance, so no equations concerning v_{y0} would work, and v_{x0} = v_0\cos{\theta_0} only works if the vertical distance is the same as the original vertical distance..

There are 2 equations of motion. Vertical and horizontal.
They work independently
They are only related by time since it is a single stone.

 

[HallsofIvy]

Doesn't seem like you need to use vectors or integration for this problem, unless I'm missing something.

Just write the two equations for the motion in the vertical and horizontal directions. The horizontal velocity stays constant (ignoring air resistance). The vertical motion follows the equations for motion given a constant acceleration (gravity).

You use the initial vertical velocity and position to figure out how long it takes the ball to hit the ground. Then use that time and the horizontal velocity to figure out how far out it hits.

Lets see some equations!

Using "vertical and horizontal directions" IS using vectors and the "equations for motion given a constant acceleration" are derived by integration. You can sweep them under the rug but they are there none the less.

 

[berkeman]

Using "vertical and horizontal directions" IS using vectors and the "equations for motion given a constant acceleration" are derived by integration. You can sweep them under the rug but they are there none the less.

True, but I was worried that mentioning that approach would scare the OP into thinking that the problem was more complicated than it is.

 

[harts]

First figure out how much time it takes for the stone to hit the ground using the fact that
y=y₀+v_ot+1/2gt²

Knowing this time, you should be able to calculate the horizontal distance traveled, using the equation you posted