Projective motion from a height, final velocity and horizontal distance

AI Thread Summary
A man throws a rock from a 10.0 m tall building at a speed of 24.0 m/s and an angle of 30.0 degrees. To find the final velocity just before impact, the correct approach involves calculating the vertical and horizontal components of velocity separately and using the equations of motion. The time of flight must be determined accurately, as the initial equations used may not apply due to the height of the launch. The conservation of energy can also be employed to find the final speed. Properly applying these principles will yield the correct final velocity and horizontal distance traveled.
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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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