Projectile motion of 2 dimensions

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a rock thrown from a 17.0 m tall building with an initial velocity of 24.0 m/s at a 33.0° angle. The correct approach involves calculating the vertical and horizontal components of the initial velocity using trigonometric functions: v0y = 24sin(33) and v0x = 24cos(33). The maximum height, final velocity before impact, and horizontal distance to the impact point can be determined using kinematic equations, specifically considering the effects of gravity and the time of flight.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of kinematic equations
  • Ability to perform trigonometric calculations
  • Familiarity with vector components of motion
NEXT STEPS
  • Calculate the vertical component of the initial velocity using v0y = 24sin(33)
  • Determine the time of flight until the rock hits the ground using kinematic equations
  • Calculate the final velocity just before impact using v = v0y + at
  • Find the horizontal distance using the horizontal component of velocity and time of flight
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone seeking to understand the calculations involved in two-dimensional motion problems.

Musicman
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A man stands on the roof of a 17.0 m tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 33.0° above the horizontal. You can ignore air resistance.

(a) Calculate the maximum height above the roof reached by the rock.

(b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

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


Ok I started out by doing 24sin(33) and 24cos(33). Then I did 17=0+23.99t+1/2(-9.8)t^2.
I got wrong answers for all 3 parts. I got 23.99 m for part A, for part B I got 44.21 m/s, and part C i got apprx 30 m. Can someone confirm my answers and not just show the answer alone, i need to know HOW to do it.
 
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Hey folks,

Just a reminder of the PF Global Guidelines that we all agreed to, under Homework Help:

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.[/color]

I've added the color for emphasis. Please don't post full solutions to problems. Helping is good. Doing the problem for the questioner is bad.

Thanks,

Tom
 
Musicman said:
(a) Calculate the maximum height above the roof reached by the rock.

What is the y-component of the velocity when the rock reaches its maximum height?

(b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

You have functions that describe both the position and the velocity of the rock as functions of time. Because you know the coordinates of the rock just before it hits the ground, you should be able to find the time at which it hits the ground. Then you should be able to find the velocity at that time.

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

You're being asked about the x-coordinate when the rock hits the ground. You should know what time that is after completing part b.

Ok I started out by doing 24sin(33) and 24cos(33). Then I did 17=0+23.99t+1/2(-9.8)t^2.

Where did the 23.99 come from? You're supposed to have [itex]v_{0y}[/itex] in that spot, but [itex]v_{0y}[/itex] certainly doesn't equal 23.99 m/s.

Try to follow the hints I gave. If you're still stuck then post what you've got and we'll try to get you un-stuck.
 

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