Projective motion from a height, final velocity and horizontal distance

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Homework Help Overview

The problem involves a rock being thrown from a height of 10.0 m with an initial velocity of 24.0 m/s at an angle of 30.0 degrees above the horizontal. The objective is to determine the magnitude of the velocity just before impact with the ground and the horizontal distance traveled.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the components of velocity and the appropriate equations to use for projectile motion from a height. There is uncertainty regarding the validity of the time of flight equation used by the original poster, with suggestions to consider energy conservation and alternative kinematic equations.

Discussion Status

Some participants have provided guidance on using different equations of motion and conservation of energy to approach the problem. There is an ongoing exploration of the correct methods to apply, with no explicit consensus reached on a single approach.

Contextual Notes

Participants note that the original poster's time of flight equation may not be applicable due to the initial height of the throw. There is also mention of needing to focus on the vertical motion separately from the horizontal motion.

benbaum
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Homework Statement


A man stands on the roof of a 10.0 m -tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 30.0 degrees above the horizontal. You can ignore air resistance.

Calculate the magnitude of the velocity of the rock just before it strikes the ground

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Homework Equations



For Velocity:

Vy (final velocity)=V0y-gT where g is 9.8 <br>

V0y (initial Y)= v0sin(theta) = 12 (?) Where V0 is initial velocity

time in flight (usable?) = 2V0y/G =2.45?

For range Delta X=Vx * T where V0x=V0Cos(theta)

Vfinal= square root of(vy^2 + vx^2)

The Attempt at a Solution



I think my real problem might be I'm not using the right equations.

Anyway, my approach was to get the two components of my final velocity by getting Vy and Vx and then use the Pythagorean theorem to get the final velocity. Initial velocity of Y and X component were easy, I plugged the given values into them and got 20.78 for V0x= V0cos(theta) and 12 for V0y=V0sin(theta). Since the initial velocity of X component is constant, I set out to solve for the rest of the final velocity for the Y component.

The first equation I used was the "time in flight" equation, which I'm not sure if it works in one of these types of problems, since it is from a height. When solving that equation I got 2.45 for T and plugged it into the final y velocity equation Vy =12-9.8*2.45 =-12.01.

Finally, I used the "Vfinal" equation square root of(20.78^2+-12.01^2) = 24.001m/s which is apparently wrong.

What am I doing wrong?
 
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Your time in flight equation seems to be derived only for when dY=0, so I believe you cannot use it. I would use V2=V02+2ay. V is the final velocity, V0 is the initial velocity, a is acceleration and y is change in height. Remember to use only velocities in the y-direction. This also works if you have an acceleration in the x-direction but no given time. Good luck!
 
Consider using a conservation of energy argument to find the final speed (magnitude of the velocity).

You'll need to develop the equation of motion for the y-direction and solve for the time of flight (time from launch until the stone hits the ground). Then you'll be in a position to find the range.
 
find velocity using conservation of energy. then you would have got acceleration, initial velocity, final velocity, horizontal and vertical velocity. simply use equation of motion
 

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